Let the centre of a circle C be [(, )]

Question

Let the centre of a circle C be $$(\alpha, \beta)$$

Accepted Answer

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

After solving, point $\mathrm{A}$ is $(4,-5)$

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

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

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

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

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

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

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

$$
\Rightarrow \text { radius } \approx 49.78 \text { (rejected })
$$

Hence, $\alpha=1, \beta=-1, \mathrm{r}=5$

$$
\alpha-\beta+\mathrm{r}=7
$$

Solutions · 1