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.

Answer 1

Let the equation of the required circle be (x – h)² + (y – k)² = r².

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

(2 – h)² + (3 – k)² = r²   …(1)

(–1 – h)² + (1 – k)² = r²   …(2)

Since the centre (h, k) of the circle lies on 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 + h² + 9 – 6k + k² = 1 + 2h + h² + 1 – 2k + k²

⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

⇒ 6h + 4k = 11   …(4)

Solving equations (3) and (4) we obtain ${\displaystyle \text{h}=\frac{7}{2}}$ and ${\displaystyle \text{k}=\frac{-5}{2}}$.

Substituting the values ​​of h and k in equation (1), we obtain

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

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

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

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

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

Thus, the equation of the required circle is

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

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

= 4x² − 28x + 49 + 4y² + 20y + 2 = 130

= 4x² + 4y² − 28x + 20y − 56 = 0

= 4(x² + y² − 7x + 5y − 14) = 0

= x² + y² − 7x + 5y − 14 = 0