Question

If (2,5) is an interior point of the circle x2 + y2 -8x -12y +P = 0 and the circle neither cuts nor touches any one of the co-ordinate axes then-
(A) P belongs to (36,47)
(b) P belongs to (16,47)
(c) P belongs to (16,36)

Answer 1

The correct answer is option (A).

$\large\underline{\underline{\text{Explanation}}}$

The general form of the equation of the circle

$\cdots \longrightarrow (x^{2}-8x+16)+(y^{2}-12y+36)+P-16-36=0$

$\cdots \longrightarrow (x-4)^{2}+(y-6)^{2}+P-52=0$

$\cdots \longrightarrow (x-4)^{2}+(y-6)^{2}=52-P.$

Let the radius be $r,$

$\cdots \longrightarrow r^{2}=52-P.$

A point lying inside the circle

Since $(2,5)$ lies inside the circle,

$\cdots \longrightarrow r^{2}>5.$

Since the circle touches neither $x=0$ nor $y=0,$ the distance from the center is less than the radius.

The perpendicular lines dropped from the center

→ Towards the x-axis has the length of 6.

→ Towards the y-axis has the length of 4.

The smaller length is 4. So, the radius is less than the length of 4.

Hence,

$\cdots \longrightarrow r^{2}<16.$

Interval notation of P

So,

$\cdots \longrightarrow 5<r^{2}<16.$

Hence,

$\cdots \longrightarrow 5<52-P<16.$

And therefore,

$\cdots \longrightarrow 36<P<47.$

According to the interval notation,

$\cdots \longrightarrow P \in (36,47).$