`P` is a point `(a , b)` in the first quadrant. If the two circles which pass through `P` and touch both the coordinates axes cut at right angles, then `a^2-6a b+b^2=0` `a^2+2a b-b^2=0` `a^2-4a b+b^2=0` `a^2-8a b+b^2=0`

Options

A. `a^(2)-6ab+b^(2)=0`
B. `a^(2)+2ab-b^(2)=0`
C. `a^(2)-4ab+b^(2)=0`
D. `a^(2)-8ab+b^(2)=0`

Answer

3

Explanation

Let the equation of the two circles be
`(x-r)^(2)+(y-r)^(2)=r^(2)`
i.e.., `x^(2)+y^(2)-2rx-2ry+r^(2)=0`, where `r=r_(1)r_(2)`
The condition of orthogonality gives
`2r_(1)r_(2)+2r_(1)r_(2)=r_(1)^(2)+r_(2)^(2)` or `4r_(1)r_(2)=r_(1)^(2)=r_(2)^(2)` (1)
Circle passes through `(a,b)`. Therefore,
`a^(2)+b^(2)-2ra-2rb+r^(2)=0`
i.e., `r^(2)-2r(a+b)+a^(2)+b^(2)=0`
`:. r_(2) +r_(2)=2(a+b)` and `r_(1)r_(2)=a^(2)+b^(2)`
`:. 4(a^(2)+b^(2))=4(a+b)^(2)-2(a^(2)+b^(2))` [From (1)]
i.e., `a^(2)-4ab+b^(2)=0`