Given two circles `x^2 +y^2+3sqrt(2)(x+y)=0` and `x^2 +y^2 +5sqrt(2)(x+y)=0`. Let the radius of the third circle, which touches the two given circles and to their common diameter, be `(2lambda-1)/lambda` The value of `lambda` is

Options

A. 10
B. 8
C. 7
D. 5

Answer

B

Explanation

Let `C_(1) and C_(2)` be the centres of the circles `x^(2)+y^(2)=3sqrt(2) (x+y)=0 and x^(2)+y^(2)+5sqrt(2) (x+y)=0` and let their radii be `r_(1) and r_(2)` respetively. Then,

`C_(1) ((-3)/(sqrt(2)), (-3)/(sqrt(2))), C_(2) ((-5)/(sqrt(2)), (-5)/(sqrt(2))), r_(1)=3 and r_(2)=5`
Let C be the centre and r be the radius of the circle touching the given circles and their common diameter at point P. Then,
`C C_(1)^(2)=CP^(2)+C_(1)P^(2)`
`rArr (r + r_(1))^(2) = r^(2)+(C_(1)C_(2)+C_(2)P)^(2)`
`rArr (r+r_(1))^(2)=r^(2)+(C_(1)C_(2)+sqrt(C C_(2)^(2)-CP^(2)))^(2)`
`rArr (r+r_(1))^(2)=r^(2)+{(r_(2)-r_(1))+sqrt((r_(2)-r)^(2)-r^(2))}^(2)`
`rArr (3+r)^(2)=r^(2)+{(2+sqrt((5-r)^(2)-r^(2))}^(2)`
`rArr 9 + 6r = (2 + sqrt(25-10r))^(2)`
`rArr 9 + 6r = 4 + 25 - 10r + 4 sqrt(25-10r)`
`rArr (4r-5)^(2)=25 - 10r rArr 16 r^(2)-30r=0 rArr r = (15)/(8)`
`:. (2 lambda-1)/(lambda) = (15)/(8) rArr lambda = 8`