Question

If $(2,6)$ is an interior point of the circle$x^{2} + y^{2} - 8x - 12y + p = 0$ and the circle neither cuts nor touches any one of the axes of co-ordinates then

• A. p∈(36,47)
• B. p ∈(16, 47)
• C. p∈(16, 36)
• D. None of these

Answer 1

Answer: (B)

p ∈(16, 47)

Answer 2

We have $x^{2} + y^{2} - 8x - 12y + p = 0$ then centre and radius of the circle are (4,6) and $\sqrt{(52 - p)}$ respectively.

$\because$Circle neither cuts nor touches any one of the axes of coordinates then

$x -$coordinates of centre > radius i.e.,

$4 > \sqrt{52 - p}$

⇒ $p > ⥂ 36$ ……………..(1)

& y-coordinate of centre > radius

$6 > \sqrt{52 - p}$,

⇒ $p > ⥂ 16$ …………….(2)

$\therefore$ D is interior point of the circle then

CD < radius $\sqrt{5} < \sqrt{52 - p}$

⇒ $p < 47$ …………(3)

from (1), (2) & (3) we obtain

$36 < p < 47$

$\therefore$p ∈(36, 47).