To determine the condition under which the line [tex]\( px + qy + r = 0 \)[/tex] is normal to the circle [tex]\( x^2 + y^2 + 2gx + 2fy + c = 0 \)[/tex], we need to consider geometric properties of circles and lines.
First, let's summarize the necessary components:
### Circle Equation:
[tex]\[
x^2 + y^2 + 2gx + 2fy + c = 0
\][/tex]
The standard form of a circle equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex] can be derived by completing the square:
- The center of the circle [tex]\((h, k)\)[/tex] is given by [tex]\((-g, -f)\)[/tex].
- The radius [tex]\(r\)[/tex] can be found using the formula:
[tex]\[
r = \sqrt{g^2 + f^2 - c}
\][/tex]
### Line Equation:
[tex]\[
px + qy + r = 0
\][/tex]
A line is normal to the circle if it passes through the circle's center because the radius at the point of tangency to a normal line is perpendicular to that line.
### Conditions:
For the line [tex]\( px + q y + r = 0 \)[/tex] to be normal to the circle, it must pass through the center of the circle.
Therefore, the coordinates of the center [tex]\((-g, -f)\)[/tex] should satisfy the line equation. Substituting [tex]\(-g\)[/tex] for [tex]\(x\)[/tex] and [tex]\(-f\)[/tex] for [tex]\(y\)[/tex] in the line equation gives us:
[tex]\[
p(-g) + q(-f) + r = 0
\][/tex]
Simplifying it, we get:
[tex]\[
-pg - qf + r = 0
\][/tex]
Which further simplifies to:
[tex]\[
pg + qf - r = 0
\][/tex]
In conclusion, the condition under which the line [tex]\( px + qy + r = 0 \)[/tex] can be normal to the circle [tex]\( x^2 + y^2 + 2gx + 2fy + c = 0 \)[/tex] is:
[tex]\[
p(-g) + q(-f) + r = 0
\][/tex]
