The two circles which pass through `(0,a)a n d(0,-a)` and touch the line `y=m x+c` will intersect each other at right angle if `a^2=c^2(2m+1)` `a^2=c^2(2+m^2)` `c^2=a^2(2+m^2)` (d) `c^2=a^2(2m+1)`

Options

A. `a^(2)=c^(2)(2m+1)`
B. `a^(2)=c^(2)(2+m^(2))`
C. `c^(2)=a^(2)(2+m^(2))`
D. `c^(2)=a^(2)(2m+1)`

Answer

3

Explanation

The equatio of the family of circles through (0,a) and (0,-a) is
`[x^(2)+(y-a)(y+a)]+lambdax=0,lambda in R`
or `x^(2)+y^(2)+lambdax-a^(2)=0`
Since circles touch line `y=mx+c`,
`sqrt(((lambda)/(2))^(2)+a^(2))=(|(-mlambda//2)+c|)/(sqrt(1+m^(2)))`
or `(1+m^(2))[(lambda^(2))/(4)+a^(2)] = ((mlambda)/(2)-c)^(2)`
or `(1+m^(2))[(lambda^(2))/(4)+a^(2)]=(m^(2)lambda^(2))/(4)-mclambda +c^(2)`
or `lambda^(2)+4mclambda +4a^(2)(1+m^(2))-4c^(2)=0`
`:. lambda_(1)lambda_(2)=4[a(1+m^(2))-c^(2)]`
or `g_(1)g_(2)=[a^(2)(1+m^(2))-c^(2)]`
Now, `g_(1)g_(2)=f_(1)f_(2)=(c_(1)+c_(2))/(2)` (As circle cut orthogonally.)
`:. a^(2)(1+m^(2))-c^(2) = -a^(2)`
Hence, `c^(2)=a^(2)(2+m^(2))`