Question

The equation of a circle passing through (3, –6) and touching both the axes is ______.

Options

• x² + y² – 6x + 6y + 8 = 0

• x² + y² + 6x – 6y + 9 = 0

• x² + y² + 30x – 30y + 225 = 0

• x² + y² – 30x + 30y + 225 = 0

Answer 1

The equation of a circle passing through (3, –6) and touching both the axes is `underlinebb(x^2 + y^2 - 30x + 30y + 225 = 0)`.

Explanation:

Circle has to pass through (3, –6) in fourth quadrant.

∴ Let the circle is (x – r)² + (g + r)² = r²

⇒ (3 – r)² + (– 6 + r)² = r²

⇒ r² – 18r + 45 = 0

⇒ r = 15, 3

∴ The circle are x² + y² – 6x + 6y + 9 = 0 or x² + y² – 30x + 30y + 225 = 0