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Q.
The equation of the circle passing through the point $(-1,-3)$ and touching the line $4 x+3 y-12=0$ at the point $(3$, 0 ) is
Conic Sections
Solution:
Family of circles touching the line $4 x+3 y-12=0$ at point $(3,0)$ is
$(x-3)^{2}+(y-0)^{2}+\lambda(4 x+3 y-12)=0$
If it passes through the point $(-1,-3)$ then
$(-1-3)^{2}+(-3-0)^{2}+\lambda(4(-1)+3(-3)-12)=0$
$\Rightarrow \lambda=1$
Hence, equation of circle is $(x-3)^{2}+(y-0)^{2}+(4 x+3 y-12)$ $=0$
or $x^{2}+y^{2}-2 x+3 y-3=0$.