If $A$ is the midpoint of the common chord of circle $x^{2} + y^{2} - 4 x - 4 y = 0$ and $x^{2} + y^{2} = 16$ and $P$ be any point on the circumference of the circle $x^{2} + 8 x + y^{2} + 12 x + 36 = 0$ then maximum length of $A P$ is

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Solution

First we need to find the points of intersection of two circles
Given
$S_{1} = x^{2} + y^{2} - 4 x - 4 y = 0 (1)$
and $S_{2} = x^{2} + y^{2} = 16 (2)$
Equation of common chord will be
$S_{1} - S_{2} = 4 x + 4 y = 16 (3)$
On solving eqn $(2)$ and $(3)$ , we get intersection points are $(4, 0)$ and $(0, 4)$ and the common chord will pass from this two points
So, the mid point of the chord will be
$x = \frac{x_{1} + x_{2}}{2}$ and $y = \frac{y_{1} + y_{2}}{2}$
The mid point $A$ is $(2, 2)$ .

Now we need to find the maximum value of $A P$ , we know that if we draw a line segment from an external point to a circle then the line segment having maximum and minimum length length will be along the normal.

Let the centre and radius of the circle $S_{3}$ be $C_{3}$ and $r_{3}$ where $S_{3} = x^{2} + 8 x + y^{2} + 12 x + 36 = 0$
Here we got $C_{3} = (- 4, - 6)$ and $r_{3} = 4$
So the distance will be
$A P = A C_{3} + r_{3}$
$A C_{3} = \sqrt{(2 - (- 4))^{2} + (2 - (- 6))^{2}}$
$A C_{3} = 10$

$∴ Length of A P = 10 + 4 = 14$

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If A is the midpoint of the common chord of circle x2+y2−4x−4y=0 and x2+y2=16 and P be any point on the circumference of the circle x2+8x+y2+12x+36=0 then maximum length of AP is

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