If the equation of the common tangent of the | vedclass

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

Question

(D) The equations of the circles are $S_1: x^2+y^2-4x+6y+4=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,the center $C_1 = (2, -3)$ and radius $r_1 = \sq

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) The equations of the circles are $S_1: x^2+y^2-4x+6y+4=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,the center $C_1 = (2, -3)$ and radius $r_1 = \sqrt{2^2+(-3)^2-4} = \sqrt{4+9-4} = 3$. For $S_2$,the center $C_2 = (-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-(-2)} = \sqrt{1+1+2} = 2$. The distance between the centers is $C_1C_2 = \sqrt{(2 - (-1))^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5$. Since $r_1 + r_2 = 3 + 2 = 5 = C_1C_2$,the circles touch each other externally at point $P$. The point of contact $P$ divides the line segment $C_1C_2$ internally in the ratio $r_1:r_2 = 3:2$. $P = \left( \frac{3(-1) + 2(2)}{3+2}, \frac{3(1) + 2(-3)}{3+2} \right) = \left( \frac{-3+4}{5}, \frac{3-6}{5} \right) = \left( \frac{1}{5}, -\frac{3}{5} \right)$. The common tangent at $P$ is the radical axis of the two circles,given by $S_1 - S_2 = 0$. $(x^2+y^2-4x+6y+4) - (x^2+y^2+2x-2y-2) = 0$ $-6x + 8y + 6 = 0$ Dividing by $-2$,we get $3x - 4y - 3 = 0$. Comparing this with $ax+by+c=0$,we have $a=3, b=-4, c=-3$. Thus,$\frac{a}{c} = \frac{3}{-3} = -1$.

If the equation of the common tangent of the circles $x^2+y^2-4x+6y+4=0$ and $x^2+y^2+2x-2y-2=0$ at their point of contact is $ax+by+c=0$,then $\frac{a}{c}=$

Similar Questions

If the lengths of the tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the coordinates of point $P$ are:

The circles $x^2 + y^2 + 2x - 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ touch each other:

The centre of a circle which cuts $x^{2}+y^{2}+6x-1=0$,$x^{2}+y^{2}-3y+2=0$ and $x^{2}+y^{2}+x+y-3=0$ orthogonally is

The equation of the radical axis of the circles $x^2 + y^2 + x - y + 2 = 0$ and $3x^2 + 3y^2 - 4x - 12 = 0$ is:

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module