Question

Find the equation to the common chord of the two circles $x^2+y^2-4x+6y-36=0$ and $x^2+y^2-5x+8y-43=0$ and also find its length.

Answer 1

Equation of common chord would be 9x-2y+7=0

Length $2\sqrt{\frac{4140}{85}}\to \approx 14$

Explanation:

Say points of intersection of two circles are $x_1,y_1$ and $x_2,y_2$

Since both the points lie on both the circles, they will satisfy the eq of both the circles. Hence

$x^2+y^2-4x+6y-36=0$ and

$x^2+y^2-5x+8y-43=0$

Subtract the 2nd eq from 1st to eliminate $x_1^2$ and $y_1^2$ to get

$9x_1-2y_1+7=0$

Like wise for the other point $x_2$ and $y_2$ we will have

$9x_2-2y_2+7=0$

These two equations show that the line 9x-2y +7=0 passes through both the points of intersection of the circles.

Hence this is the equation of the common chord.

The first circle eq in standard form would be $(x+2{)}^2+(y+3{)}^2=7^2$

The centre of the circle is at (-2, -3). Length of the perpendicular from this point on to the line 9x-2y+7=0 would be $\frac{|9(-2)-2(-3)+7|}{\sqrt{9^2+(-2{)}^2}}=\frac{5}{\sqrt{85}}$

Since line joining the two centres is always perpendicular to common chord and bisects the common chord, a right triangle will be formed of which the hypotenuse would be the radius of the circle (in this case it is 7) and the adjacent sides would be $\frac{5}{\sqrt{85}}$ and half length of the common chord.

Using Pythagorus theorem this half length would be $\sqrt{49-(\frac{5}{\sqrt{85}}{)}^2}=\sqrt{\frac{4140}{85}}$

The length of the common chord would be twice of this $=2\sqrt{\frac{4140}{85}}$