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 `2sqrt(4140/85)-> 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_1 ^2 +y_1 ^2 +4x_1 +6y_1-36=0` and

`x_1 ^2 +y_1^2 -5x_1 +8y_1 -45=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 pint `x_2, 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 `|9(-2) -2(-3) +7|/sqrt (9^2 +(-2)^2) =5/ sqrt85`
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 `5/sqrt85` and half length of the common chord.

Using Pythagorus theorem this half length would be `sqrt (49-(5/sqrt85)^2`= `sqrt(4140/85)`

The length of the common chord would be twice of this =`2sqrt(4140/85)`