The length of the common chord of the two circles `x^2+y^2-4y=0` and `x^2+y^2-8x-4y+11=0` is

Options

A. `(sqrt(145))/(4)` cm
B. `(sqrt(11))/(2) cm `
C. `sqrt(135)` cm
D. `(sqrt(135))/(4) cm`

Answer

D

Explanation

Given , equation of cirlcesare
`x^(2) + y^(2) - 4y = 0`
and ` x^(2) + y^(2) - 8x - 4y + 11=0`
`therefore ` Equation of chord is
`x^(2) + y^(2) - 4y -(x^(2) + y^(2)- 8x - 4y + 11) =0`
`rArr" "8x 11=0`

So , centre and radius of first circle are O(0,2) and OP = r= 2.
Now , perpendicular distance form O(0,2) to the line 8x - 11 is
`d = OM = (| 8 xx 0 - 11|)/(sqrt(8^(2))) = (11)/(8)`
In `DeltaOMP ,PM = sqrt(OP^(2) - OM^(2))`
`= sqrt(2^(2) -((11)/(8))^(2)) = sqrt(4-(121)/(64))`
` = sqrt(256-121)/(64) = sqrt(135)/(4) cm `