Question

A circle passes through the origin and has its centre on the line y = x. If it cuts x² + y² – 4x – 6y + 10 = 0 orthogonally, equation of the circle is ______.

पर्याय

• x² + y² – y – x = 0

• x² + y² – 2x – 2y = 0

• x² + y² + 2x + 2y = 0

• x² + y² – 6x – 4y = 0

Answer 1

A circle passes through the origin and has its centre on the line y = x. If it cuts x² + y² – 4x – 6y + 10 = 0 orthogonally, equation of the circle is x² + y² – 2x – 2y = 0.

Explanation:

Let the equation of the circle be

x² + y² + 2gx + 2fy + c = 0  ...(i)

∴ Coordinates of centre of the circle = (– g, – f)

As the circle passes through the origin,

0² + 0² + 2g(0) + 2f(0) + c = 0

`\implies` c = 0

Given, the centre lies on y = x

`\implies` Coordintes of the centre are (– g, – g).

Given, two circles, x² + y² + 2gx + 2fy + c = 0 and

x² + y² – 4x – 6y + 10 = 0 are orthogonal.

Therefore, 2 × g × (–2) + 2 × f × (–3) = c + 10  ...[∵ 2g₁g₂ + 2f₁f₂ = c₁ + c₂]

`\implies` –4g – 6f = c + 10

`\implies` –10g = c + 10   ...[∵ g = f]

`\implies` –10g = 10    ...[∵ c = 0]

`\implies` g = – 1 

∴  f = – 1

Hence, the equation of circle is x² + y² – 2x – 2y = 0.