Question

If the line 3x – 4y – k = 0, (k > 0) touches the circle

x² + y² – 4x – 8y – 5 = 0 at (a, b) then k + a + b is equal to-

• A. 20
• B. 22
• C. – 30
• D. –28

Answer 1

Answer: (A)

20

Answer 2

Since the given line touches the given circle, the length of the perpendicular from the centre (2, 4) of the circle to the line 3x – 4y – k = 0 is equal to the radius $\sqrt{4 + 16 + 5}$ = 5 of the circle.

Ž $\frac{3 \times 2 - 4 \times 4–k}{\sqrt{9 + 16}}$= ± 5 Ž k = 15 [Q k > 0]

Now equation of the tangent at (a, b) to the given circle is

xa + yb – 2 (x + a) – 4(y + b) – 5 = 0

Ž (a – 2)x + (b – 4)y – (2a + 4b + 5) = 0.

If it represents the given line 3x – 4y – k = 0

then $\frac{a - 2}{3}$ = $\frac{b - 4}{- 4}$ = $\frac{2a + 4b + 5}{k}$ = l (say)

then a = 3l + 2, b = 4 – 4l and 2a + 4b + 5 = kl (1)

Ž 2(3l + 2) + 4(4 – 4l) + 5 = 15l (Q k = 15)

Ž l = 1 Ž a = 5, b = 0 and k + a + b = 20.