Prove that the straight lines joining the origin to the point of intersection of the straight line `h x+k y=2h k` and the curve `(x-k)^2+(y-h)^2=c^2` are perpendicular to each other if `h^2+k^2=c^2dot`

Options

A. `h^2+k2=c^2`
B. `h^2+k^2=2c^2`
C. `h^2-k^2=c^2`
D. None of these

Answer

A

Explanation

We have , `kx+hy=2hk`.....(i)
and `x^2+y^2-2(hx+ky)+h^2+k^2-c^2=0` ......(ii)
The combined equation of straight lines joining the origin to the points of intersection of Eqs. (i) and (ii) is `x^2+y^2-2(hx+ky)((kx+hy)/(2hk))+(h^2+k^2-c^2)((ky+hy)/(2hk))^2=0`
`rArr (h^2y^2+k^2h^2)(h^2+k^2-c^2)-2hk(h^2+k^2+c^2)xy=0`
Lines given by this equation are at right angle.
`therefore (k^2+h^2)(h^2+k^2-c^2)=0`
`rArr h^2+k^2=c^2`