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Jump to heading Module 7–03 Circle

Jump to heading 1.Equation of a Circle

Jump to heading $1$ Standard Form

A circle with center $(x_{0}, y_{0})$ and radius $r$ can be represented by the equation: $(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} .$
Equation derivations
$\begin{array}{ll} \sqrt{(x - x_{0})^{2} + (y - y_{0})^{2}} = r & Two-point distance formula \\ (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} \\ Example: \\ (x + 3)^{2} + (y - 2)^{2} = 4 \\ Center: (- 3, 2) & Letting r = 0 makes the circle shrink to a point — the center \\ Radius: r = 2 & 2^{2} = 4 \end{array}$

Jump to heading $2$ General Form

$x^{2} + y^{2} + a x + b y + c = 0.$
Jump to heading It can be completed into the standard form: $(x + \frac{a}{2})^{2} + (y + \frac{b}{2})^{2} = \frac{a^{2} + b^{2} - 4 c}{4} .$
- Center: $(- \frac{a}{2}, - \frac{b}{2}) .$
- Radius: $r = \frac{a^{2} + b^{2} - 4 c}{2} .$
Special cases:
- $a = 0 : x^{2} + y^{2} + b y + c = 0.$ Center on the y-axis.
- $b = 0 : x^{2} + y^{2} + a x + c = 0.$ Center on the x-axis.
- $c = 0 : x^{2} + y^{2} + a x + b y + c = 0.$ The function passes through the origin.
Jump to heading Note: The condition for the general form to represent a circle is $a^{2} + b^{2} - 4 c > 0.$
Equation derivations
$\begin{array}{ll} (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} & Standard form \\ x^{2} - 2 x_{0} x + x_{0}^{2} + y^{2} - 2 y_{0} y + y_{0}^{2} = r^{2} & Expand the perfect square \\ x^{2} + y^{2} - 2 x_{0} x - 2 y_{0} y + (x_{0}^{2} + y_{0}^{2}) = r^{2} \\ x^{2} + y^{2} - 2 x_{0} x - 2 y_{0} y + (x_{0}^{2} + y_{0}^{2} - r^{2}) = 0 \\ Let a = - 2 x_{0} b = - 2 y_{0} c = x_{0}^{2} + y_{0}^{2} - r^{2} \\ x^{2} + y^{2} + a x + b y + c = 0 \end{array}$

Jump to heading 2.Special Circles (Standard Form)

Special Circles	Equations	Properties
$x_{0} = 0$	$x^{2} + (y - y_{0})^{2} = r^{2}$	Center on the y-axis
$y_{0} = 0$	$(x - x_{0})^{2} + y^{2} = r^{2}$	Center on the x-axis
$x_{0} = y_{0} = 0$	$x^{2} + y^{2} = r^{2}$	Center at the origin
$\| y_{0} \| = r$	$(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$	Tangent to the x-axis
$\| x_{0} \| = r$	$(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$	Tangent to the y-axis
$\| x_{0} \| = \| y_{0} \| = r$	$(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$	Tangent to both axes

Jump to heading 3.Focus 1

Equation of a Circle

Pay attention to the requirements of the circle's equation, as well as the forms of semicircle equations.

Jump to heading $22$ Given that $x^{2} + y^{2} - 4 x + 6 y + m = 0$ represents a circle, what is the range of values for $m ? .$

$\begin{array}{lllll} (A) m < 12 & (B) m < 13 & (C) m \leq 12 & (D) m \leq 13 & (E) m > 13 \end{array}$

Jump to heading Solution

$1$ Solve by completing the square to convert the general form into the standard form
$\begin{array}{ll} (x + \frac{a}{2})^{2} + (y + \frac{b}{2})^{2} = \frac{a^{2} + b^{2} - 4 c}{4} \\ (x + \frac{- 4}{2})^{2} + (y + \frac{6}{2})^{2} = \frac{(- 4)^{2} + 6^{2} - 4 m}{4} \\ (x - 2)^{2} + (y + 3)^{2} = \frac{16 + 36 - 4 m}{4} = \frac{52 - 4 m}{4} \\ (x - 2)^{2} + (y + 3)^{2} = 13 - m > 0 & Radius > 0 for a valid circle. \\ m < 13 \end{array}$

$2$ Solve using the condition for the general form of a circle.
$\begin{array}{ll} x^{2} + y^{2} + a x + b y + c = 0 \\ a^{2} + b^{2} - 4 c > 0 & Necessary condition for a circle. \\ (- 4)^{2} + 6^{2} - 4 m > 0 \\ 16 + 36 - 4 m > 0 \\ 52 - 4 m > 0 \\ \frac{52}{4} > m \\ m < 13 \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $m < 13$ , so choose $B$ .
Formula used
$\begin{array}{ll} (x + \frac{a}{2})^{2} + (y + \frac{b}{2})^{2} = \frac{a^{2} + b^{2} - 4 c}{4} & Complete to standard form \\ x^{2} + y^{2} + a x + b y + c = 0 & General form of a circle \\ a^{2} + b^{2} - 4 c > 0 & General form circle condition \end{array}$

Jump to heading $23$ If the equation of a circle is $x^{2} + y^{2} = 1$ , then what is the equation of its right semicircle (the part located in the first and fourth quadrants) $? .$

$\begin{array}{lllll} (A) y - \sqrt{1 - x^{2}} = 0 & (B) x - \sqrt{1 - y^{2}} = 0 & (C) y + \sqrt{1 - x^{2}} = 0 \\ (D) x + \sqrt{1 - y^{2}} = 0 & (E) x^{2} + y^{2} = \frac{1}{2} \end{array}$

Jump to heading Solution

Show known conditions
$\begin{array}{ll} x^{2} + y^{2} = 1 \\ x^{2} = 1 - y^{2} \\ x = \sqrt{1 - y^{2}} & x > 0 since it's the right half. \\ x - \sqrt{1 - y^{2}} = 0 & right semicircle \\ y - \sqrt{1 - x^{2}} = 0 \to y = \sqrt{1 - x^{2}} & y \in R^{+} upper semicircle \\ y + \sqrt{1 - x^{2}} = 0 \to y = - \sqrt{1 - x^{2}} & y \in R^{-} lower semicircle \\ x + \sqrt{1 - y^{2}} = 0 \to x = - \sqrt{1 - y^{2}} & x \in R^{-} left semicircle \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $x - \sqrt{1 - y^{2}} = 0$ , so choose $B$ .
Additionally, if the problem is a standard form equation of a circle
$\begin{array}{ll} (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} \\ (x - x_{0})^{2} = r^{2} - (y - y_{0})^{2} \\ x = x_{0} + \sqrt{r^{2} - (y - y_{0})^{2}} & left semicircle \\ x = x_{0} - \sqrt{r^{2} - (y - y_{0})^{2}} & right semicircle \\ (y - y_{0})^{2} = r^{2} - (x - x_{0})^{2} \\ y = y_{0} + \sqrt{r^{2} - (x - x_{0})^{2}} & upper semicircle \\ y = y_{0} - \sqrt{r^{2} - (x - x_{0})^{2}} & lower semicircle \end{array}$

Jump to heading 4.Focus 2

Intersection of a circle and the coordinate axes

Let $y = 0$ to find the points where the circle intersects the x-axis; let $x = 0$ to find the points where it intersects the y-axis. If the circle has only one point of intersection with an axis, then it is tangent to that axis.

Jump to heading $24$ What are the two points where the circle $x^{2} + (y - 1)^{2} = 4$ intersects the $x -$ axis $? .$

$\begin{array}{lllll} (A) (- \sqrt{5}, 0), (\sqrt{5}, 0) & (B) (- 2, 0), (2, 0) \\ (C) (0, - \sqrt{5}), (0, \sqrt{5}) & (D) (- \sqrt{3}, 0), (\sqrt{3}, 0) \\ (E) (- \sqrt{2}, - \sqrt{3}), (\sqrt{2}, \sqrt{3}) \end{array}$

Jump to heading Solution

$\begin{array}{ll} Let y = 0 \\ x^{2} + (0 - 1)^{2} = 4 \\ x^{2} + 1 = 4 \\ x = \pm \sqrt{3} \Rightarrow (- \sqrt{3}, 0), (\sqrt{3}, 0) \end{array}$

Jump to heading Conclusion

Derived Solution
$(D)$
According to the Solution, get $x = \pm \sqrt{3} \Rightarrow (- \sqrt{3}, 0), (\sqrt{3}, 0)$ , so choose $D$ .

Jump to heading $25$ What is the equation of the circle centered at $P (- 2, 3)$ and tangent to the $y -$ axis $? .$

$\begin{array}{lllll} (A) (x - 2)^{2} + (y + 3)^{2} = 4 & (B) (x + 2)^{2} + (y - 3)^{2} = 4 \\ (C) (x - 2) + (y + 3)^{2} = 9 & (D) (x + 2)^{2} + (y - 3)^{2} = 9 \\ (E) (x - 3)^{2} + (y + 2)^{2} = 9 \end{array}$

Jump to heading Solution

$\begin{array}{ll} | x_{0} | = r & Tangent to the y -axis \\ r = | - 2 | = 2 \\ ((- 2) - x_{0})^{2} + (3 - y_{0})^{2} = 2^{2} \\ (x + 2)^{2} + (y - 3)^{2} = 4 \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $(x + 2)^{2} + (y - 3)^{2} = 4$ , so choose $B$ .
Formula used
$\begin{array}{ll} | x_{0} | = r & Special circles in standard form \\ (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} & Standard form of a circle \end{array}$

Jump to heading 5.Position of a Point Relative to a Circle

Let $P (x_{p}, y_{p})$ be a point, and let the circle be defined by $(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$ ,
Substitute the point into the circle's equation:
$(x_{p} - x_{0})^{2} + (y_{p} - y_{0})^{2} {\begin{cases} < r^{2} the point lies inside the circle. \\ = r^{2} the point lies on the circle. \\ > r^{2} the point lies outside the circle. \end{cases}$

Jump to heading 6.Relationship Between a Line and a Circle

Given the line $l : y = k x + b$ and the circle $O : (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$ , let $d$ be the distance from the center of the circle $(x_{0}, y_{0})$ to the line $l .$

Line–Circle Position Relationship	Diagram	Condition (Geometric Interpretation)
Line and circle are separate No Intersection		$d > r$
Line tangent to circle 1 Intersection Point		$d = r$
Line intersects circle 2 Intersection Points		$d < r$

Chord length of a circle
- Derived from the Pythagorean theorem.
- $Chord length = 2 \sqrt{r^{2} - d^{2}} .$

Jump to heading 7.Relationship Between Two Circles

Let $O_{1} : (x - x_{1})^{2} + (y - y_{1})^{2} = r_{1}^{2},$ and $O_{2} : (x - x_{2})^{2} + (y - y_{2})^{2} = r_{2}^{2},$ where we may assume $r_{1} > r_{2} .$ Let $d$ be the distance between the centers $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) .$

Circle–Circle Position Relationship	Condition (Geometric Interpretation)	Number of Common Internal Tangents	Number of Common External Tangents
Externally separate No Intersection	$d > r_{1} + r_{2}$	2	2
Externally tangent 1 Intersection Point	$d = r_{1} + r_{2}$	1	2
Intersecting 2 Intersection Points	$\| r_{1} - r_{2} \| < d < r_{1} + r_{2}$	0	2
Internally tangent 1 Intersection Point	$d = \| r_{1} - r_{2} \|$	0	1
Internally contained No Intersection	$d < \| r_{1} - r_{2} \|$	0	0

The range of the distance $d$ between the circles and their position relationship.

Jump to heading 8.Focus 3

The positional relationship between a point and a circle

First, substitute the point into the equation of the circle, then make the judgment.

Jump to heading $26$ If the point $P (2 m, m)$ is inside the circle $x^{2} + y^{2} - 4 x + 2 y + 1 = 0$ , what is the range of values for $m ? .$

$\begin{array}{lllll} (A) \frac{1}{5} < m < 1 & (B) - \frac{1}{5} < m < 1 & (C) m < \frac{1}{5} \lor m > 1 \\ (D) - 1 < m < \frac{1}{5} & (E) - 1 < m < - \frac{1}{5} \end{array}$

Jump to heading Solution

$\begin{array}{ll} The point is inside the circle \Rightarrow< r^{2} \\ P (2 m, m) \\ 4 m^{2} + m^{2} - 8 m + 2 m + 1 < 0 & Substitute P \\ 5 m^{2} - 6 m + 1 < 0 \\ 5 m^{2} - 6 m + 1 \Rightarrow (1 m - 1) (5 m - 1) = 0 \\ m = 1 \lor m = \frac{1}{5} \\ Parabola Analysis: \\ 5 > 0 : Opens upward \\ Roots : Intersects m -axis at m = 1 and m = \frac{1}{5} \\ Range : f (x) < 0 between the two roots \\ f (x) < 0 ⟶ \frac{1}{5} < m < 1 \end{array}$

Jump to heading Conclusion

Derived Solution
$(A)$
According to the Solution, get $f (x) < 0 ⟶ \frac{1}{5} < m < 1$ , so choose $A$ .
Formula used
$\begin{array}{ll} (x_{p} - x_{0})^{2} + (y_{p} - y_{0})^{2} {\begin{cases} < r^{2} the point lies inside the circle. \\ = r^{2} the point lies on the circle. \\ > r^{2} the point lies outside the circle. \end{cases} & Point-circle relationship \end{array}$

Jump to heading 9.Focus 4

The positional relationship between a line and a circle

First, find the distance d from the center of the circle to the line. Then compare the sizes of d and r to determine their relationship. The most important positional relationship is tangency. Additionally, when the line intersects the circle, you should be able to use the Pythagorean theorem to find the chord length: $Chord length = 2 \sqrt{r^{2} - d^{2}} .$

Jump to heading $27$ The line $y = k (x + 2)$ is a tangent to the circle $x^{2} + y^{2} = 1.$ What is the value of $k ? .$

$\begin{array}{lllll} (A) \pm \frac{\sqrt{3}}{2} & (B) \frac{\sqrt{3}}{3} & (C) - \frac{\sqrt{3}}{3} & (D) \pm \frac{\sqrt{3}}{3} & (E) \pm \sqrt{3} \end{array}$

Jump to heading Solution

$1$ Solve by using the geometric method with $r = d$ to find the intersection point
$\begin{array}{ll} To line's general form \to point-to-line distance formula. \\ y = k (x + 2) \\ y = k x + k 2 \\ 0 = k x - y + k 2 \\ From center (0, 0) \to k x - y + k 2 = 0 \\ d = \frac{| 2 k |}{\sqrt{k^{2} + 1^{2}}} = 1 & r=1 \\ | 2 k | = \sqrt{k^{2} + 1} \\ (2 k)^{2} = (\sqrt{k^{2} + 1})^{2} \\ 4 k^{2} = k^{2} + 1 \\ 3 k^{2} = 1 \\ k^{2} = \frac{1}{3} \\ k = \pm \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \\ k = \pm \frac{\sqrt{3}}{3} \end{array}$

$2$ Solve by using the algebraic method with simultaneous equations to find the intersection point
$\begin{array}{ll} {\begin{cases} y = k (x + 2) \\ x^{2} + y^{2} = 1 \end{cases} \\ x^{2} + k^{2} (x + 2)^{2} = 1 \\ x^{2} + k^{2} (x^{2} + 4 x + 4) = 1 \\ x^{2} + k^{2} x^{2} + 4 k^{2} x + 4 k^{2} = 1 \\ (1 + k^{2}) x^{2} + 4 k^{2} x + (4 k^{2} - 1) = 0 \\ Δ = 0 & Tangent \\ Δ = (4 k^{2})^{2} - 4 (1 + k^{2}) (4 k^{2} - 1) = 0 \\ 16 k^{4} - 4 (1 + k^{2}) (4 k^{2} - 1) = 0 \\ 16 k^{4} - 4 [(1) (4 k^{2} - 1) + k^{2} (4 k^{2} - 1)] = 0 \\ 16 k^{4} - 4 [4 k^{2} - 1 + 4 k^{4} - k^{2}] = 0 \\ 16 k^{4} - 4 [3 k^{2} + 4 k^{4} - 1] = 0 \\ 16 k^{4} - [12 k^{2} + 16 k^{4} - 4] = 0 \\ 16 k^{4} - 16 k^{4} - 12 k^{2} + 4 = 0 \\ - 12 k^{2} + 4 = 0 \\ k^{2} = \frac{- 4}{- 12} = \frac{1}{3} \\ k = \pm \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \\ k = \pm \frac{\sqrt{3}}{3} \end{array}$

$3$ Solve by using graphical analysis to find the intersection point (only applicable in simple or special cases)
$\begin{array}{ll} y = k (x + 2) \\ y = y_{0} + k (x_{0} + 2) \\ \overset{Let both sides = 0}{\overset{⏞}{y_{0} = 0 ⟹ the x-intercept of the line is (–2, 0)}} \\ r=1 & (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} \\ 1 : \sqrt{3} : 2 = 30^{\circ}, 60^{\circ}, 90^{\circ} & Special right triangle \\ 30^{\circ} = \frac{\sqrt{3}}{3} & Angle–slope reference \\ k = \pm \frac{\sqrt{3}}{3} \end{array}$

Jump to heading Conclusion

Derived Solution
$(D)$
According to the Solution, get $k = \pm \frac{\sqrt{3}}{3}$ , so choose $D$ .
Formula used
$\begin{array}{ll} d = r & Line circle relationship \\ d = \frac{| c |}{\sqrt{a^{2} + b^{2}}} & Point-to-line distance formula \\ a x + b y + c = 0 & General form of a line \\ y = y_{0} + k (x - x_{0}) & Point-slope form \\ Δ = b^{2} - 4 a c {\begin{cases} > 0 : Two distinct real roots → intersecting \\ = 0 : One real root (a repeated root) → tangent \\ < 0 : No real roots → separate \end{cases} & Quadratic discriminant \\ {\begin{cases} (a + b)^{2} = a^{2} + 2 a b + b^{2} \\ (a - b)^{2} = a^{2} - 2 a b + b^{2} \end{cases} & Perfect square formula \\ 30^{\circ} = \frac{\sqrt{3}}{3} & Inclination angle–slope reference \\ 1 : \sqrt{3} : 2 & Special right triangle: 30-60-90 \\ (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} & Standard form of a circle \end{array}$

Jump to heading $28$ Given that the center of circle $C$ is the intersection point of the line $x - y + 1 = 0$ and the x-axis, and that circle $C$ is tangent to the line $x + y + 3 = 0$ , what is the equation of circle $C ? .$

$\begin{array}{lllll} (A) (x - 1)^{2} + y^{2} = 2 & (B) (x + 1)^{2} + y^{2} = 2 \\ (C) (x + 1)^{2} + y^{2} = 4 & (D) x^{2} + (y + 1)^{2} = 2 \\ (E) x^{2} + (y - 1)^{2} = 2 \end{array}$

Jump to heading Solution

$\begin{array}{ll} x - y + 1 = 0 \\ y = x + 1 & Slope-intercept form \\ \overset{Let both sides = 0}{\overset{⏞}{y = 0 ⟹ the x-intercept of the line is (–1, 0)}} & Center \\ x + y + 3 = 0 \\ d = \frac{| - 1 + 3 |}{\sqrt{1 + 1}} = \frac{2 \sqrt{2}}{2} = \sqrt{2} & Substitute (- 1, 0) \\ r = \sqrt{2} \\ (x - (- 1))^{2} + (y - 0)^{2} = {\sqrt{2}}^{2} \\ (x + 1)^{2} + y^{2} = 2 \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $(x + 1)^{2} + y^{2} = 2$ , so choose $B$ .
Formula used
$\begin{array}{ll} y = k x + b & Slope-intercept form \\ d = \frac{| a x_{0} + b y_{0} + c |}{\sqrt{a^{2} + b^{2}}} & Point-to-line distance formula \\ (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} & Standard form of a circle \end{array}$

Jump to heading🔗 Module 7–03 Circle

Jump to heading🔗 1.Equation of a Circle

Jump to heading🔗 1Standard Form

Jump to heading🔗 2General Form

Jump to heading🔗 It can be completed into the standard form: (x+a2)2+(y+b2)2=a2+b2−4c4.

Jump to heading🔗 Note: The condition for the general form to represent a circle is a2+b2−4c>0.

Jump to heading🔗 2.Special Circles (Standard Form)

Jump to heading🔗 3.Focus 1

Jump to heading🔗 22Given that x2+y2−4x+6y+m=0 represents a circle, what is the range of values for m?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 23If the equation of a circle is x2+y2=1, then what is the equation of its right semicircle (the part located in the first and fourth quadrants)?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 4.Focus 2

Jump to heading🔗 24What are the two points where the circle x2+(y−1)2=4 intersects the x-axis ?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 25What is the equation of the circle centered at P(−2,3) and tangent to the y-axis?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 5.Position of a Point Relative to a Circle

Jump to heading🔗 6.Relationship Between a Line and a Circle

Jump to heading🔗 7.Relationship Between Two Circles

Jump to heading🔗 8.Focus 3

Jump to heading🔗 26If the point P(2m,m) is inside the circle x2+y2−4x+2y+1=0, what is the range of values for m?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 9.Focus 4

Jump to heading🔗 27The line y=k(x+2) is a tangent to the circle x2+y2=1. What is the value of k?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 28Given that the center of circle C is the intersection point of the line x−y+1=0 and the x-axis, and that circle C is tangent to the line x+y+3=0, what is the equation of circle C?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading Module 7–03 Circle

Jump to heading 1.Equation of a Circle

Jump to heading $1$ Standard Form

Jump to heading $2$ General Form

Jump to heading It can be completed into the standard form: $(x + \frac{a}{2})^{2} + (y + \frac{b}{2})^{2} = \frac{a^{2} + b^{2} - 4 c}{4} .$

Jump to heading Note: The condition for the general form to represent a circle is $a^{2} + b^{2} - 4 c > 0.$

Jump to heading 2.Special Circles (Standard Form)

Jump to heading 3.Focus 1

Jump to heading $22$ Given that $x^{2} + y^{2} - 4 x + 6 y + m = 0$ represents a circle, what is the range of values for $m ? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $23$ If the equation of a circle is $x^{2} + y^{2} = 1$ , then what is the equation of its right semicircle (the part located in the first and fourth quadrants) $? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 4.Focus 2

Jump to heading $24$ What are the two points where the circle $x^{2} + (y - 1)^{2} = 4$ intersects the $x -$ axis $? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $25$ What is the equation of the circle centered at $P (- 2, 3)$ and tangent to the $y -$ axis $? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 5.Position of a Point Relative to a Circle

Jump to heading 6.Relationship Between a Line and a Circle

Jump to heading 7.Relationship Between Two Circles

Jump to heading 8.Focus 3

Jump to heading $26$ If the point $P (2 m, m)$ is inside the circle $x^{2} + y^{2} - 4 x + 2 y + 1 = 0$ , what is the range of values for $m ? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 9.Focus 4

Jump to heading $27$ The line $y = k (x + 2)$ is a tangent to the circle $x^{2} + y^{2} = 1.$ What is the value of $k ? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $28$ Given that the center of circle $C$ is the intersection point of the line $x - y + 1 = 0$ and the x-axis, and that circle $C$ is tangent to the line $x + y + 3 = 0$ , what is the equation of circle $C ? .$

Jump to heading Solution

Jump to heading Conclusion