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)+k^(2)=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=2hx" …(i)"`
and, `x^(2)+y^(2)-2(hx+ky)+h^(2)+k^(2)-c^(2)=0" (ii)"`
The combined equation of the straight lines joining the origin to the points of intersection of (i) and (ii) is
`x^(2)+y^(2)-2(hx+ky)((kx+hy)/(2hk))+(h^(2)+k^(2)-c^(2))((kx+hy)/(2hk))^(2)=0`
`rArr" "(h^(2)y^(2)+k^(2)x^(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))=0rArrh^(2)+k^(2)=c^(2)`