Question

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

Answer 1

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 $(-1, 1)$
$(2 - h)^2 + (3 - k)^2 = r^2 … (1)$
$(-1 - h)^2 + (1 - k)^2 = r^2 … (2)$
Since the center (h, k) of the circle lies on the line
$x - 3y - 11 = 0,$
$h - 3k = 11 … (3)$
From equations (1) and (2), we obtain
$(2 - h)^2 + (3 - k)^2 = (-1 - h)^2 + (1 - k) ^2$
$⇒ 4 - 4h + h ^2 + 9 - 6k + k^2 = 1 + 2h + h ^2 + 1 - 2k + k ^2$
$⇒ 4 - 4h + 9 - 6k = 1 + 2h + 1 - 2k$
$⇒ 6h + 4k = 11 … (4)$
On solving equations (3) and (4), we obtain $h = 7/2$ and $k = -5/2$
On substituting the values of h and k in equation (1), we obtain
$(2 –\frac{ 7}{2})^2 \+ (3 + \frac{5}{2})^2 = r^2$
[$[\frac{(4-7)}{2}^2] \+ [\frac{(6+5)}{2}^2] = r^2$
$(-\frac{3}{2})^2 \+ (\frac{11}{2})^2 = r^2$
\frac{9}{4} + \frac{121}{4} = r^2$$$ \frac{130}{4} = r^2$Thus, the equation of the required circle is$(x – \frac{7}{2})^2+ (y + \frac{5}{2})^2= \frac{130}{4}$$4x^2-28x + 49 +4y^2+ 20y + 25 =130$$4x^2 +4y^2-28x + 20y – 56 = 0$$4(x^2+y^2-7x + 5y – 14) = 0$$x^2 + y^2 – 7x + 5y – 14 = 0$∴ The equation of the required circle is$x^2 + y^2 – 7x + 5y – 14 = 0$