(Each question 4 marks)

Question

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

Accepted Answer

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 $(-1, 1), (2 - h)^2 + (3 - k)^2 = r^2 ..... (i) (-1 - h)^2 + (1 - k)^2 = r^2 .....$ (ii) Since the centre (h, k) of the circle lies on line $x - 3y - 11 = 0, h - 3k = 11 ....$. (iii) From equations (i) and (ii),
we obtain $(2 - h)^2+ (3 - k)^2 = (-1 - h)^2 + (1 - k)^2$
$\Rightarrow 4 - 4h + h^2 + 9 - 6k + k^2 = 1 + 2h + h^2 + 1 - 2k + k^2 $
$\Rightarrow 4 - 4h + 9 - 6k = 1 + 2h + 1 - 2k $
$\Rightarrow 6h + 4k = 11 … $(iv) On solving equations (iii) and (iv), we obtain $\text{h}=\frac{7}{2}$ and $\text{k}=\frac{-5}{2}$ On substituting the values of h and k in equation (i), we obtain $\Big(\text{x}-\frac{7}{2}\Big)^2+\Big(\text{y}+\frac{5}{2}\Big)^2=\frac{130}{4}$ $\Big(\frac{2\text{x}-7}{2}\Big)^2+\Big(\frac{2\text{y}+5}{2}\Big)^2=\frac{130}{4}$ $4\text{x}^2-28\text{x}+49+4\text{y}^2+20\text{y}+25=130$ $4\text{x}^2+4\text{y}^2-28\text{x}+20\text{y}-56=0$ $4(\text{x}^2+\text{y}^2-7\text{x}+5\text{y}-14)=0$ $\text{x}^2+\text{y}^2-7\text{x}+5\text{y}-14=0$

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