The combined equation of the direct common ta | vedclass

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

Question

(A) For the circle $x^2+y^2+2x=0$,the center is $C_1(-1,0)$ and radius $r_1=1$.For the circle $x^2+y^2-2y-3=0$,the center is $C_2(0,1)$ and radius $r_

Vedclass · Accepted Answer

$xy+x+2y+2=0$

Vedclass · Answer

(A) For the circle $x^2+y^2+2x=0$,the center is $C_1(-1,0)$ and radius $r_1=1$.For the circle $x^2+y^2-2y-3=0$,the center is $C_2(0,1)$ and radius $r_2=2$.The distance between centers is $C_1C_2 = \sqrt{(0-(-1))^2 + (1-0)^2} = \sqrt{1^2+1^2} = \sqrt{2}$.Since $r_2 - r_1 = 2 - 1 = 1 Wait,the direct common tangents meet at the external center of similitude $P$,which divides $C_1C_2$ externally in the ratio $r_1:r_2 = 1:2$.$P = \left(\frac{1(0)-2(-1)}{1-2}, \frac{1(1)-2(0)}{1-2}\right) = \left(\frac{2}{-1}, \frac{1}{-1}\right) = (-2,-1)$.The lines passing through $P(-2,-1)$ with slope $m$ are $y+1 = m(x+2) \Rightarrow mx-y+2m-1=0$.The distance from $C_1(-1,0)$ to this line is $r_1=1$: $\frac{|m(-1)-0+2m-1|}{\sqrt{m^2+(-1)^2}} = 1 \Rightarrow |m-1| = \sqrt{m^2+1}$.Squaring both sides: $m^2-2m+1 = m^2+1$ $\Rightarrow -2m=0$ $\Rightarrow m=0$.If $m=0$,the line is $y+1=0$. The other tangent is a vertical line passing through $P(-2,-1)$,which is $x+2=0$.The combined equation is $(y+1)(x+2) = 0 \Rightarrow xy+x+2y+2=0$.

The combined equation of the direct common tangents of the circles $x^2+y^2+2x=0$ and $x^2+y^2-2y-3=0$ is

Similar Questions

If $m_1$ and $m_2$ are the slopes of the direct common tangents drawn to the circles $x^2+y^2-2x-8y+8=0$ and $x^2+y^2-8x+15=0$,then $m_1+m_2=$

Find the equation of the circle passing through the intersection of the circle $x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $3x + 4y + 5 = 0$,and also passing through the point $(1, 2)$.

The equation of a circle which passes through the points of intersection of the circles $2x^2+2y^2-2x+6y-3=0$ and $x^2+y^2+4x+2y+1=0$,and whose centre lies on the common chord of these circles is

The distance of the origin from the external centre of similitude for the circles $x^2+y^2-8x-10y-8=0$ and $x^2+y^2+2x-2y-2=0$ is

The number of common tangents to the two circles $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module