Question

Equation of tangent having slope $1$ to the circle $x^2+y^2-10x-8y+5=0$ is

• A. $x-y+1-5\sqrt{2}=0$
• B. $x-y-1-6\sqrt{2}=0$
• C. $x-y+1-6\sqrt{2}=0$
• D. $x-y-1+\sqrt{2}=0$

Answer 1

The correct option is B $x-y-1-6\sqrt{2}=0$

Given circle equation $x^2+y^2-10x-8y+5=0$ and slope of tangent is $1$

centre of circle $=(-g,-f)=(5,4)$; radius $=\sqrt{g^2+f^2-c}=\sqrt{25+16-5}=\sqrt{36}=6$

let equation of tangent $y=x+c(m=1)$

Distance of centre to tangent $=r$

$\left|\frac{4-5-c}{\sqrt{2}}\right|=6\Rightarrow |c+1|=6\sqrt{2}\Rightarrow c=6\sqrt{2}-1,-6\sqrt{2}-1$

$\Rightarrow y=x+6\sqrt{2}-1$ or $y=x-6\sqrt{2}-1$