Let the centre of a circle C be ( , ) and its radius r

Question

Let the centre of a circle C be $(\alpha ,\beta )$ and its radius $r<8$. Let $3x+4y=24$ and $3x-4y=32$ be two tangents and $4x+3y=1$ be a normal to C. Then $(\alpha -\beta +r)$ is equal to :

Accepted Answer

First find point A by solving $4x+3y=1$ and $3x-4y=32$

After solving, point $A$ is $(4,-5)$

Centre $(\alpha ,\beta )$ lie on $4x+3y=1$

$4\alpha +3\beta =1\Rightarrow \beta =\frac{1-4\alpha }{3}$

Now distance from centre to line $3x-4y-32=$ 0 and $3x+4y-24=0$ are equal.

$|\frac{3\alpha -4(\frac{1-4\alpha }{3})-32}{5}|=|\frac{3\alpha +4(\frac{1-4\alpha }{3})-24}{5}|$

After solving $\alpha =1$ and $\alpha =\frac{28}{3}$

For $\alpha =1$, centre $(1,-1)\Rightarrow$ radius $=5$

For $\alpha =\frac{28}{3}$, centre $(\frac{28}{3},\frac{-109}{2})$

$\Rightarrow \approx 49.78)$

Hence, $\alpha =1,\beta =-1,r=5$

$\alpha -\beta +r=7$

Solutions · 1