Question

If the circle $C_1:x^2+y^2=16$ intersects another circle $C_2$ of radius 5 in such a manner that the common chord of maximum length $8$ has a slope equal to $\frac{3}{4}$, then coordinates of centre of $C_2$ are -

• A. $(\frac{9}{5},-\frac{12}{5})$
• B. $(-\frac{9}{5},\frac{12}{5})$
• C. $(\frac{9}{5},\frac{12}{5})$
• D. $(-\frac{9}{5},-\frac{12}{5})$

Answer 1

The correct option is A $(\frac{9}{5},-\frac{12}{5})$

The maximum length of chord = Diameter of circle $C_1=8units$

Now, equation of the chord of slope $\frac{3}{4}$,

$3x-4y=0$.

The center of circle $C_2$ must be on the line perpendicular to the chord.

So, the center of the circle $C_2$ can be written as $(3a,-4a)$

$x_1=3a,y_1=-4a$

Where $a$ common factor.

As the chord shown in image, we get $AB=5$ units using pythagoras theorem.

Hence,

Using sectional formula from a point to a line,

$\frac{3\times \left(3a\right)}{5}-\frac{4\times (-4a)}{5}=8-5$

$25a=15$

$=>a=\frac{3}{5}$.

Now,

The center of the circle $C_2$ can be written as $(3\times \frac{3}{5},-4\times \frac{3}{5})=(\frac{9}{5},\frac{-12}{5})$.