Question

1. find the length of the tangent from the point (5,7) to the circle $x^{2}+y^{2}-4x - 6y+9 = 0$

Explanation:

Step1: Rewrite the circle equation in standard form

The general equation of a circle is $x^{2}+y^{2}+2gx + 2fy + c=0$, and its standard - form is $(x + g)^{2}+(y + f)^{2}=r^{2}$, where the center of the circle is $(-g,-f)$ and the radius $r=\sqrt{g^{2}+f^{2}-c}$.
For the circle equation $x^{2}+y^{2}-4x - 6y + 9 = 0$, we have $2g=-4$, $2f=-6$, $c = 9$. So $g=-2$, $f=-3$.
The center of the circle $C$ is $(2,3)$ and the radius $r=\sqrt{(-2)^{2}+(-3)^{2}-9}=\sqrt{4 + 9-9}=2$.

Step2: Calculate the distance between the point and the center of the circle

The distance $d$ between the point $P(5,7)$ and the center $C(2,3)$ of the circle is given by the distance formula $d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}$.
Here, $x_1 = 2,y_1 = 3,x_2 = 5,y_2 = 7$. So $d=\sqrt{(5 - 2)^{2}+(7 - 3)^{2}}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step3: Use the Pythagorean theorem to find the length of the tangent

Let the length of the tangent be $l$. By the Pythagorean theorem, $l=\sqrt{d^{2}-r^{2}}$.
Since $d = 5$ and $r = 2$, then $l=\sqrt{25-4}=\sqrt{21}$.

Answer:

$\sqrt{21}$