The equation of the circle whose radius is 3 | vedclass

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

Question

(A) The given circle is $S: x^2+y^2-4x-6y-12=0$. The center $C_1$ is $(2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$. Let the req

Vedclass · Accepted Answer

$5x^2+5y^2-8x-14y-32=0$

Vedclass · Answer

(A) The given circle is $S: x^2+y^2-4x-6y-12=0$. The center $C_1$ is $(2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$. Let the required circle have center $C_2(h, k)$ and radius $r_2 = 3$. Since the circles touch internally at $P(-1, -1)$,the point $P$ divides the line segment $C_1C_2$ externally in the ratio $r_1 : r_2 = 5 : 3$. Using the section formula for external division: $-1 = \frac{5h - 3(2)}{5-3} \implies -2 = 5h - 6 \implies 5h = 4 \implies h = \frac{4}{5}$. $-1 = \frac{5k - 3(3)}{5-3} \implies -2 = 5k - 9 \implies 5k = 7 \implies k = \frac{7}{5}$. The equation of the circle is $(x - \frac{4}{5})^2 + (y - \frac{7}{5})^2 = 3^2$. $x^2 - \frac{8}{5}x + \frac{16}{25} + y^2 - \frac{14}{5}y + \frac{49}{25} = 9$. $x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y + \frac{65}{25} - 9 = 0$. $x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y + \frac{13}{5} - 9 = 0$. $x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y - \frac{32}{5} = 0$. Multiplying by $5$,we get $5x^2 + 5y^2 - 8x - 14y - 32 = 0$.

The equation of the circle whose radius is $3$ and which touches the circle $x^2+y^2-4x-6y-12=0$ internally at $(-1,-1)$ is

Similar Questions

Let the centre of a circle,passing through the points $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$,be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k)$,$4(h^2+k^2)$ is equal to .............

If the circles $x^2+y^2-2x-2(3+\sqrt{7})y+8+6\sqrt{7}=0$ and $x^2+y^2-8x-6y+k^2=0, k \in \mathbb{Z}$,have exactly two common tangents,then the number of possible values of $k$ is

Consider the family of circles $x^2+y^2-2x-2\lambda y-8=0$. This family passes through two fixed points $A$ and $B$. Find the distance between these two points.

If two circles touch at $(0,0)$ externally,then their radical axis is

If the radical centre of the three circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$,and $x^2+y^2-2y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles,then the equation of the circle with $C(\alpha, \beta)$ as centre and $r$ as radius is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module