# Thread: equation of a circle!

1. ## equation of a circle!

ok so i got the tangent lines of the two points,

point A(4,3) y=4/3x-25/3

point B(4,-3) y=-4/3x+25/3

now i just need help finding the normal lines...thanks...

mathlete

The normal line to a curve at a point is perpendicular to the tangent line at that point.
Find the equation of the normal line to this circle at Point A.
y(x) =
Find the equation of the normal line to this circle at Point B.
y(x) =

(d) At what point do the two normal lines intersect?

2. Originally Posted by mathlete
ok so i got the tangent lines of the two points,
point A(4,3) y=4/3x-25/3

point B(4,-3) y=-4/3x+25/3
Are they
$\displaystyle A(4,3)~$, $\displaystyle ~y=\frac{25}{3} - \frac{4x}{3}$
and
$\displaystyle A(4,-3)~$, $\displaystyle ~y=\frac{4x}{3}-\frac{25}{3}$
?

First,
$\displaystyle A(4,3)~$
$\displaystyle ~y=\frac{25}{3} - \frac{4x}{3}$
The line has the slope $\displaystyle -\frac{4}{3}$.
So the line perpendicular to it should have the slope $\displaystyle \frac{3}{4}$

The general formula is,
$\displaystyle y = mx + C$, m is the slope, C is a constant
$\displaystyle y = \frac{3}{4}x + C$
It must pass $\displaystyle A(4,3)$
$\displaystyle 3 = \frac{3}{4}4 + C$
$\displaystyle C = 0$
So the equation of the normal is,
$\displaystyle y = \frac{3}{4}x$

You can apply the same for the other line.