# For the circle, (x+1)^2+(y+2)^2=25 What is the length of the tangent to the circle from the point (6,4)?

Mar 25, 2018

The length of the tangent to the circle from the point $\left(6 , 4\right)$ to circle ${\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = 25$ is $2 \sqrt{15}$

#### Explanation:

The circle ${\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = 25$ is a circle with center $\left(- 1 , - 2\right)$ and radius $5$.

As the length of the tangent from the external point, radius to the point at which tangent touches circle and line joining external point to center of circle form a right angle.

Now radius is $\sqrt{25} = 5$, distance between external point and center of circle is $\sqrt{{\left(6 + 1\right)}^{2} + {\left(4 + 2\right)}^{2}}$

and hence using Pythagorus theorem, lengt of tangent is $\sqrt{{\left(6 + 1\right)}^{2} + {\left(4 + 2\right)}^{2} - 25} = \sqrt{49 + 36 - 25} = \sqrt{60} = 2 \sqrt{15}$

graph{((x+1)^2+(y+2)^2-0.03)((x+1)^2+(y+2)^2-25)((x-6)^2+(y-4)^2-0.03)=0 [-10, 10, -4.8, 5.2]}

Obsserve that the length of the tangent to a circle ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$ from an external point $\left({x}_{1} , {y}_{1}\right)$ will always be $\sqrt{{x}_{1}^{2} + {y}_{1}^{2} + 2 g {x}_{1} + 2 f {y}_{1} + c}$.