Question

Equation of a circle having radius equal to twice the radius of the circle x² + y² + (2p + 3)x + (3 – 2p)y + p – 3 = 0 and touching it at the origin is –

• A. x² + y² + 9x – 3y = 0
• B. x² + y² – 9x + 3y = 0
• C. x² + y² + 18x + 6y = 0
• D. x² + y² + 18x – 6y = 0

Answer 1

Answer: (D)

x² + y² + 18x – 6y = 0

Answer 2

Since the given circle passes through the origin p – 3 = 0 Ž p = 3 and the equation of the given circle is

x² + y² + 9x – 3y = 0

Equation of the tangent at the origin to this circle is

9x – 3y = 0 (i)

Let the equation of the required circle which also passes through the origin be x² + y² + 2gx + 2ƒy = 0.

Equation of the tangent at the origin to this circle is

gx + ƒy = 0 (ii)

If (i) and (ii) represent the same line, then

$\frac{g}{9} = \frac{ƒ}{–3}$= k (say) (iii)

We are given that $\sqrt{g^{2} + ƒ^{2}}$ = 2

$\sqrt{\left( \frac{9}{2} \right)^{2} + \left( \frac{- 3}{2} \right)^{2}}$= $\sqrt{81 + 9}$

From (iii) we get |k| $\sqrt{9^{2} + 3^{2}}$ = $\sqrt{90}$Ž k = ±1

For k = 1,

g = 9, ƒ = –3 and the equation of the required circle is

x² + y² + 18x – 6y = 0.