# Find the equation of the circle passing through the points (2, 3)

Question:

Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line – 3y – 11 = 0.

Solution:

Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Since the circle passes through points (2, 3) and (–1, 1),

$(2-h)^{2}+(3-k)^{2}=r^{2} \ldots(1)$

$(-1-h)^{2}+(1-k)^{2}=r^{2} \ldots(2)$

Since the centre (h, k) of the circle lies on line – 3y – 11 = 0,

– 3k = 11 … (3)

From equations (1) and (2), we obtain

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

$\Rightarrow 4-4 h+h^{2}+9-6 k+k^{2}=1+2 h+h^{2}+1-2 k+k^{2}$

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

$\Rightarrow 6 h+4 k=11 \ldots(4)$

On solving equations (3) and (4), we obtain $h=\frac{7}{2}$ and $k=\frac{-5}{2}$.

On substituting the values of $h$ and $k$ in equation (1), we obtain

$\left(2-\frac{7}{2}\right)^{2}+\left(3+\frac{5}{2}\right)^{2}=r^{2}$

$\Rightarrow\left(\frac{4-7}{2}\right)^{2}+\left(\frac{6+5}{2}\right)^{2}=r^{2}$

$\Rightarrow\left(\frac{-3}{2}\right)^{2}+\left(\frac{11}{2}\right)^{2}=r^{2}$

$\Rightarrow \frac{9}{4}+\frac{121}{4}=r^{2}$

$\Rightarrow \frac{130}{4}=r^{2}$

Thus, the equation of the required circle is

$\left(x-\frac{7}{2}\right)^{2}+\left(y+\frac{5}{2}\right)^{2}=\frac{130}{4}$

$\left(\frac{2 x-7}{2}\right)^{2}+\left(\frac{2 y+5}{2}\right)^{2}=\frac{130}{4}$

$4 x^{2}-28 x+49+4 y^{2}+20 y+25=130$

$4 x^{2}+4 y^{2}-28 x+20 y-56=0$

$4\left(x^{2}+y^{2}-7 x+5 y-14\right)=0$

$x^{2}+y^{2}-7 x+5 y-14=0$