Mathematics - Circle Question with Solution | TestHub

Lines $y=x+2$, $4y=3x+6$ and $3y=4x+1$ are tangents to the circle $(x-h{)}^2+(y-k{)}^2=r^2$.

Centre of the circle is $\left(h,k\right)$.

Equation of bisector of lines $4y=3x+6, 3y=4x+1$ is:

$\frac{4x-3y+1}{5}=\pm \left(\frac{3x-4y+6}{5}\right)$

$\Rightarrow 4x-3y+1=\pm \left(3x-4y+6\right)$

Taking positive sign, we get

$4x-3y+1=3x-4y+6$

$\Rightarrow x+y=5$

Since, centre $\left(h,k\right)$ lies on the bisector, therefore

$h+k=5$