Answer:
[tex](x-2)^{2} +(y+4)^{2} = \frac{4}{5}[/tex]
Step-by-step explanation:
Given that the center of the circle is: (2, -4)
tangent to the line x = -2
The equation of a circle has the following formula:
[tex](x-a)^{2} +(y-b)^{2} = r^{2}[/tex] where a, b is the center and r is the radius
The distance between the circle and the tangent line is:
d = [tex]\frac{2*1 + 2}{\sqrt{2^{2}+4^{2}} }[/tex] = [tex]2\frac{\sqrt{5} }{5}[/tex]
d = r so [tex]d^{2} =r^{2}[/tex] = [tex](2\frac{\sqrt{5} }{5})^{2}[/tex] = [tex]\frac{4}{5}[/tex]
the equation of the circle is:
[tex](x-2)^{2} +(y+4)^{2} = \frac{4}{5}[/tex]
