Finding Equation of a Circle given the center and tangent

Question

Our teacher told us to have a self-paced learning on pre-cal. Here's the question: Find an equation of a circle tangent to the line $x=5$ and with center $(-2,-5)$. I don't understand here is where would be the intersection at one point and find the equation.

Answer 1

Does this figure help you visualize the problem?

Answer 2

welcome to MSE.

Hint: As Matti suggested draw as below:

1- Draw $y=x$, it crosses the origin O.

2- draw a perpendicular from point $C(-2, -5)$ and mark it' foot as A.

3- use the fact that tangent on a circle is perpendicular on radius of the circle. So the distance between C and A is the radius of the circle.

4- find the equation of the perpendicular:

$y-(-5) =-1[x-(-2)]$

Or:

$y=-x-7$

Now the intersection of this line with line $y=x$ give the coordinates of A as: $A(-\frac72, -\frac72)$

and the radius of circle R is:

$R=\sqrt {(-2-7/2)^2+(-5+7/2)^2}=\frac32\sqrt2$

And equation of circle is:

$(x+2)^2+(y+5)^2=R^2=\frac92$