Question

Find the equation of a circle concentric with the circle x² + y² – 6x + 12y + 15 = 0 and has double of its area.

Answer 1

Given equation of the circle is x² + y² – 6x + 12y + 15 = 0  ......(i)

Centre = `(-g, -f) = (3, 6)`   .......`[(because 2g = -6 ⇒ g = - 3),(2f = 12 ⇒ f = 6)]`

Since the circle is concentric with the given circle

∴ Centre = (3, – 6)

Now let the radius of the circle is r

∴ r = `sqrt(g^2 + f^2 - c)`

= `sqrt(9 + 36 - 15)`

= `sqrt(30)`

Area of the given circle (i) = πr² = 30π sq.unit

Area of the required circle = 2 × 30π = 60π sq.unit

If r₁ be the radius of the required circle

πr₁² = 60π

⇒ r₁² = 60

So, the required equations of the circle is

(x – 3)² + (y + 6)² = 60

⇒ x² + 9 – 6x + y² + 36 + 12y – 60 = 0

⇒ x² + y² – 6x + 12y – 15 = 0

Hence, the required equation is x² + y² – 6x + 12y – 15 = 0.