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

The radius of the circle which cuts all the three circles $x^2+y^2-4x-4y+3=0$,$x^2+y^2+4x-4y+3=0$,and $x^2+y^2+4x+4y+3=0$ orthogonally is

The length of the tangent drawn from any point on the circle $x^2+y^2+2gx+2fy+c_1=0$ to the circle $x^2+y^2+2gx+2fy+c_2=0$ is

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is:

The equation of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $2{x^2} + 2{y^2} - 10x - 12y + 12 = 0$ is:

If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ intersect orthogonally,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module