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 $${(x-h)}^2+{(y-k)}^2=a^2$$

Given:
h = 1, k = 2

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

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

∴$${(4-1)}^2+{(6-2)}^2=a^2$$

$$\Rightarrow 9+16=a^2$$
$$\Rightarrow a=5(\because a>0)$$

Substituting the value of a in equation (1):

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

$$\Rightarrow x^2+1-2x+y^2+4-4y=25$$
$$\Rightarrow x^2-2x+y^2-4y=20$$
$$\Rightarrow x^2+y^2-2x-4y-20=0$$

Thus, the required equation of the circle is

$$x^2+y^2-2x-4y-20=0$$