Question

Find the equation to the circle which passes through the points (1, - 2) and (4, - 3) and which has its centre on the straight line 3x + 4y = 7.

Answer 1

Let the center of the circle be $(h,k)$
Since the center is equidistant from all the points on the circle,

$(h-1{)}^2+(k+2{)}^2=(h-4{)}^2+(k+3{)}^2$

$\Rightarrow h^2-2h+1+k^2+4k+4=h^2-8h+16+k^2+6k+9$

$\Rightarrow 6h-2k-20=0$ or $3h-k-10=0$ ........(1)

The center also satisfies $3h+4k-7=0$

Finding the intersection of these lines, we get

$3h-k-10$ = $3h+4k-7$

Put value in equation (1)

$5k+3=0$ or $k=\frac{-3}{5}$

$\therefore 3h=10-\frac{3}{5}=\frac{47}{5}$ or $h=\frac{47}{15}$

The radius will be $\sqrt{{(\frac{47}{15}-1)}^2+{(\frac{-3}{5}+2)}^2}=\sqrt{{\left(\frac{32}{15}\right)}^2+{\left(\frac{7}{5}\right)}^2}=\sqrt{\frac{1024+441}{225}}=\sqrt{\frac{1465}{225}}$

The circle equation is therefore ${(x-\frac{47}{15})}^2+{(y+\frac{3}{5})}^2=\frac{293}{45}$