Question

state the necessary and sufficient condition for Lx+my+n=0 to be a normal to the circle x²+y²+2gx+2fy+c=0​

Answer 1

Let's solve for c.

$\rm x^2+y^2+2gx+2fy+c=0$

Step 1: Add -2fy to both sides.

$\rm 2fy+2gx+x^2+y^2+c+−2fy=0+−2fy$

$\rm 2gx+x^2+y^2+c=−2fy$

Step 2: Add -2gx to both sides.

$\rm^2gx+x^2+y^2+c+−2gx=−2fy+−2gx$

$\rm x^2+y^2+c=−2fy−2gx$

Step 3: Add -x^2 to both sides.

$\rm x^2+y^2+c+−x^2=−2fy−2gx+−x^2$

$\rm y^2+c=−2fy−2gx−x^2$

Step 4: Add -y^2 to both sides.

$\rm y^2+c+−y^2=−2fy−2gx−x^2+−y^2$

$\rm c=−2fy−2gx−x^2−y^2$

Answer:

$\rm c=−2fy−2gx−x^2−y^2$