Question

The range of parameter ‘a’ for which the variable line

y = 2x + a lies between the circle

x² + y² – 2x – 2y + 1 = 0 and x² + y² – 16x – 2y + 61 = 0 without intersecting or touching either circle, is –

• A. (–15 + 2$\sqrt{5}$, –$\sqrt{5}$ – 1)
• B. (15 + 2$\sqrt{5}$, $\sqrt{5}$ – 1)
• C. (–15 – 2$\sqrt{5}$, –$\sqrt{5}$ + 1
• D. (– 15 + 2$\sqrt{5}$, $\sqrt{5}$ – 1)

Answer 1

Answer: (A)

(–15 + 2$\sqrt{5}$, –$\sqrt{5}$ – 1)

Answer 2

The equations of the given circles are

x² + y² – 2x – 2y + 1 = 0 … (1)

and, x² + y² – 16x – 2y + 61 = 0 … (2)

The coordinates of the centres and radii of these two circle are

C₁(1, 1), r₁ = 1 and C₂(8, 1), r₂ = 2 respectively.

For the line y = 2x + a not to touch or intersect circle (1), we must have

$\left| \frac{1 + a}{\sqrt{5}} \right|$ > 1 [Length of perpendicular from centre

C₁ > radius r₁]

Ž |a + 1| > $\sqrt{5}$

Ž a Ī (–, –$\sqrt{5}$ – 1) Č ($\sqrt{5}$ – 1, ) …(3)

Similarly, for the line y = 2x + a not to touch or intersect circle (2), we must have

$\left| \frac{15 + a}{\sqrt{5}} \right|$ > 2

Ž | 15 + a | > $2\sqrt{5}$

Ž a Ī (–, –15 – 2$\sqrt{5}$, –15 + 2$\sqrt{5}$) …(4)

The line y = 2x + a will be between the circles, if their centres C₁ and C₂ are on the opposite sides of it.

\ (2 – 1 + a) (16 – 1 + a) < 0

Ž (a + 1) (a + 15) < 0

Ž a Ī (–15, –1) … (5)

From equations (3), (4) and (5), we get

a Ī (–15 +$2\sqrt{5}$, –$\sqrt{5}$ – 1).