Question

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Answer 1

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

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x – h)² + y² = 25.

It is given that the circle passes through point (2, 3).

∴ (2 − h)² + 3² = 25

⇒ (2 − h)² = 25 − 9

⇒ (2 − h)² = 16

⇒ 2 − h = `±sqrt16` = ±4

If 2 − h = 4, then h = −2

If 2 − h = −4, then h = 6

When h = −2 the equation of the circle becomes

(x + 2)² + y² = 25

x² + 4x + 4 + y² = 25

x² + y² + 4x − 21 = 0

When h = 6 the equation of the circle becomes

(x − 6)² + y² = 25

x² − 12x + 36 + y² = 25

x² + y² − 12x + 11 = 0