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  4. The line 3 x+5 y+9=0 w.r.t. the circle x2+y2-4 x+6 y+5=0 is

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Q. The line $3 x+5 y+9=0$ w.r.t. the circle $x^2+y^2-4 x+6 y+5=0$ is

Conic Sections

Solution:

From centre $(2,-3)$, length of perpendicular on line $3 x+5 y+9=0$ is
$p=\frac{6-15+9}{\sqrt{25+9}}=0$; line is diameter.