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**arze** Form the equation of straight lines joining the origin to the points of intersection of $\displaystyle ax^2+2hxy+by^2+2gx+2fy+c=0$ and $\displaystyle px+qy+r=0$ and write down the condition that these lines should be right angles.

If this condition is satisfied, show that the equation of the locus of the foot of the perpendicular from the origin to the line $\displaystyle px+qy+r=0$ is

$\displaystyle (a+b)(x^2+y^2)+2gx+2fy+c=0$

I've done the first part. The answers I found are:

$\displaystyle x^2(ar^2-2gpr+p^2c)+2xy(hr^2-fpr-gqr+pqc)+y^2(br^2-2fqr+q^2c)=0$

and

$\displaystyle r^2(a+b)-2r(gp+fq)+c(p^2+q^2)=0$

How do I proceed with the next part?