Question

Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

Answer 1

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

Given:
h = 1, k = 2

∴ Equation of the circle = ${\left(x-1\right)}^2+{\left(y-2\right)}^2=a^2$ ...(1)

Also, equation (1) passes through (4, 6).

∴ ${\left(4-1\right)}^2+{\left(6-2\right)}^2=a^2$
$$\begin{gathered} \Rightarrow 9+16=a^2 \\ \Rightarrow a=5\left(\because a>0\right) \end{gathered}$$

Substituting the value of a in equation (1):

${\left(x-1\right)}^2+{\left(y-2\right)}^2=25$

$$\begin{gathered} \Rightarrow x^2+1-2x+y^2+4-4y=25 \\ \Rightarrow x^2-2x+y^2-4y=20 \\ \Rightarrow x^2+y^2-2x-4y-20=0 \end{gathered}$$

Thus, the required equation of the circle is $x^2+y^2-2x-4y-20=0$.