If P(2,8) is an interior point of a circle x2+y2−2x+4y−p=0 which

Discussion Forum : Circles

Question -

If $P (2, 8)$ is an interior point of a circle $x^{2} + y^{2} - 2 x + 4 y - p = 0$ which neither touches nor intersects the axes, then set for $p$ is -

Options:

A .

p < - 1

B .

p < - 4

C .

p > 96

D .

p \in ϕ

Answer: Option D
:
D
Since the point P is interior to the circle,

S_{1}

< 0

= (2^{2}) + (8^{2}) - 2. (2) + 4 (8) - p < 0

= 96 - p < 0

= p > 96

Also given that the circle doesn't touches any of the axes.
So,

g^{2} - c

< 0

f^{2} - c

< 0

g^{2} - c

< 0

= 1 + p < 0

= p < - 1

Also,

f^{2} - c

< 0

= 4 + p < 0

= p < - 4

Since

p < - 4

and

p > 96

is not possible, the set for p will be a null set meaning it'll not have any element in it.

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