# Finding an equation of a circle with a given center and a tangent line.

My math homework is finding an equation of the circle. Given that the center is at (-3,-5) and tangent to the line 12x + 5y =4.

I don't know how to solve this since our professor didn't teach this to us. :/

• If you know the center, sounds like all you need is the radius. Do you know how to find the distance between a point and a line? – Mike May 4 '14 at 8:36
• Radius = Perpendicular distance from the center to the tangent – lab bhattacharjee May 4 '14 at 8:38

Distance from the center $(-3,-5)=(x_0,y_0)$ to the tangent line $12x + 5y -4=0,A=12,B=5,C=-4$ is radius $$r=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}=65/13=5$$ equation of circle with center $(-3,-5)$ and radius $r=5$ is $$(x+3)^2+(x+5)^2=5^2$$
• Equation of a distance from a line $Ax+By+C=0$ to the point $(x_0,y_0)$ is $d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ – Adi Dani May 4 '14 at 9:26
Let $(x+3)^2 + (y+5)^2 = r^2$ be the equation of the circle with center $C = (-3,-5)$ and radius equal to $r$. Rewrite the tangent line's equation as: $y + 5 = \dfrac{29 - 12x}{5}$, substituting this $y$ into the equation of the circle and get: $(x+3)^2 + \dfrac{(29 -12x)^2}{25} - r^2 = 0 \iff 169x^2 - 546x + 1066 - 25r^2 = 0$. We require that this equation has one and only one real root. This means: $\triangle' = 0 \iff (-273)^2 - 169(1066 - 25r^2) = 0 \iff 4225r^2 - 105625 = 0 \iff r = 5$. So the equation of the circle is: $(x+3)^2 + (y+5)^2 = 25$