Question

For what value of c, the lines which join the origin to the points of intersection of the given straight line and the curve be at right angle?

For what value of c, the lines which join the origin to the points of intersection of straight line `x-y+c=0` and the curve `x^2+y^2+4x-6y-36=0` be at right angle?

Answer 1

`c=5pmsqrt97`

Explanation:

Equation of the straight line perpendicular to the sraight line `x-y+c=0` and passing through the origin `(0,0)` will be `x+y=0`.

Let us solve for `x and y` to find out the coordinates of point of intersection of the straight line `x+y=0` and the circle `x^2+y^2+4x-6y-36=0`
Inserting `x=-y` in the equation of the circle we get

`y^2+y^2-4y-6y-36=0`

`=>2y^2-10y-36=0`

`=>y^2-5y-18=0`

`=>y=(5pmsqrt(5^2-4xx1xx(-18)))/2`

`=>y=(5pmsqrt97)/2`

when `y=(5+sqrt97)/2`,then `x=-y=-(5+sqrt97)/2`

Now from given equation `x-y+c=0`

we get `c=y-x=5+sqrt97`

and

when `y=(5-sqrt97)/2`,then `x=-y=-(5-sqrt97)/2`

for this coordinate

`c=y-x=5-sqrt97`