Question

find the length of common chord of the circle
x^2+y^2-2ax-4ay-4a^2=0
and
x^2+y^2-3ax+4ay=0
​

Answer 1

Answer:

$8a\sqrt{\frac{14}{65}}\text{ unit}$

Step-by-step explanation:

Given circles,

$x^2+y^2-2ax-4ay-4a^2=0-----(1)$,

$x^2+y^2-3ax+4ay=0-------(2)$

The equation of common chord is,

Equation (1) - equation (2) = 0,

$-2ax+3ax-4ay-4ay-4a^2=0$

$ax-8ay-4a^2=0$

$x-8y-4a=0------(3)$

Equation (1) can be written as,

$(x-a)^2+(y-2a)^2-a^2-4a^2-4a^2=0$

$(x-a)^2+(y-2a)^2=9a^2$

So, the center of the circle (1) is (a, 2a),

Also, the radius of the circle = 3a,

Now, the length of perpendicular from (a, 2a) to the chord (3),

$d=|\frac{a-8(2a)-4a}{\sqrt{1+(8)^2}}|$

$=\frac{19a}{\sqrt{65}}\text{ unit}$

$\because \sqrt{(3a)^2-(\frac{19a}{\sqrt{65}})^2}$

$=\sqrt{9a^2-\frac{361a^2}{65}$

$=\sqrt{\frac{585a^2-361a^2}{65}}$

$=\sqrt{\frac{224a^2}{65}}$

$=4a\sqrt{\frac{14}{65}}$

Hence, the length of the common chord = $2\times 4a\sqrt{\frac{14}{65}}=8a\sqrt{\frac{14}{65}}\text{ unit}$