Question

Find the equation of the circle passing through the points (1, -2) and (4, -3) and whose centre lies on the line 3x + 4y = 7.

Answer 1

Let (h, k) be the centre of circle.

Since, centre of circle lies on the line 3x + 4y = 7.

$\therefore 3h+4k=7...\left(i\right)$

Since, this circle passes through the points A(1, -2) and B(4, -3),

$\therefore$ OA = OB [radii of circle]

$\Rightarrow O{A}^2=O{B}^2$

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

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

$\Rightarrow 6h-2k=20$

$\therefore 3h-k=10...\left(ii\right)$

On solving Eqs. (i) and (ii), we get

$h=\frac{47}{15}$ and $k=-\frac{3}{5}$

So, the coordinates of centre of circle $(\frac{47}{15},-\frac{3}{5}).$

$\therefore$ Radius of circle = OA

= $\sqrt{{(1-\frac{47}{15})}^2+{(-2+\frac{3}{5})}^2}=\sqrt{{(-\frac{32}{15})}^2+{\left(\frac{-7}{5}\right)}^2}$

= $\sqrt{\frac{1024}{225}+\frac{49}{25}}=\sqrt{\frac{1024+441}{225}}=\frac{\sqrt{1465}}{15}$

So, the equation of circle having centre $(\frac{47}{15},-\frac{3}{5})$ and radius $\frac{\sqrt{1465}}{15}$ is

${(x-\frac{47}{15})}^2+{(y+\frac{3}{5})}^2={\left(\frac{\sqrt{1465}}{15}\right)}^2$

$\therefore {(x-\frac{47}{15})}^2+{(y+\frac{3}{5})}^2=\frac{1465}{225}$