Find the equation of a circle with radius 3 t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the given circle be $S: x^{2} + y^{2} - 4x - 6y - 12 = 0$. Its center $C$ is $(2, 3)$ and radius $R = \sqrt{2^{2} + 3^{2} - (-12)} = \sqrt{4 +

Vedclass · Accepted Answer

$(x - \frac{4}{5})^{2} + (y - \frac{7}{5})^{2} = 3^{2}$

Vedclass · Answer

(A) Let the given circle be $S: x^{2} + y^{2} - 4x - 6y - 12 = 0$. Its center $C$ is $(2, 3)$ and radius $R = \sqrt{2^{2} + 3^{2} - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$. Let the required circle have center $C_1(h, k)$ and radius $r = 3$. Since the circle touches internally at $P(-1, -1)$,the center $C_1$ lies on the line segment $CP$. The distance $CC_1 = R - r = 5 - 3 = 2$. Thus,$C_1$ divides the line segment $CP$ in the ratio $r : (R - r) = 3 : 2$ internally. Using the section formula,$C_1 = (\frac{3(2) + 2(-1)}{3+2}, \frac{3(3) + 2(-1)}{3+2}) = (\frac{6-2}{5}, \frac{9-2}{5}) = (\frac{4}{5}, \frac{7}{5})$. The equation of the circle is $(x - \frac{4}{5})^{2} + (y - \frac{7}{5})^{2} = 3^{2}$.

Find the equation of a circle with radius $3$ that touches the circle $x^{2} + y^{2} - 4x - 6y - 12 = 0$ internally at the point $(-1, -1)$.

Similar Questions

If the circle $S \equiv x^2+y^2+2gx+4y+1=0$ bisects the circumference of the circle $x^2+y^2-2x-3=0$,then the radius of circle $S=0$ is

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

If the circles $x^2+y^2-4x+2fy+1=0$ and $x^2+y^2+2gx-4y-1=0$ cut orthogonally,then $r_1^2+r_2^2-8=$

The length of the diameter of the circle which cuts the following three circles orthogonally is:
$x^{2}+y^{2}-x-y-14=0$
$x^{2}+y^{2}+3x-5y-10=0$
$x^{2}+y^{2}-2x+3y-27=0$

Find the equation of a circle which cuts the circle $x^2+y^2-6x+4y-3=0$ orthogonally,while passing through $(3,0)$ and touching the $Y$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module