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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. :/

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  • $\begingroup$ 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? $\endgroup$ – Mike May 4 '14 at 8:36
  • $\begingroup$ Radius = Perpendicular distance from the center to the tangent $\endgroup$ – lab bhattacharjee May 4 '14 at 8:38
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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 $$(x+3)^2+(x+5)^2=5^2$$

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  • $\begingroup$ How did you come up with the solution of finding the radius? I swear to God our professor didn't teach this stuffs that's why I don't have the knowledge with this one. :/ $\endgroup$ – Z'K May 4 '14 at 9:10
  • $\begingroup$ Equation of a distance from a line $Ax+By+C=0$ to the point $(x_0,y_0)$ is $d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ $\endgroup$ – Adi Dani May 4 '14 at 9:26
  • $\begingroup$ Ah that explains. Thank you very much sir now I understand it somehow :) You are a good teacher sir. $\endgroup$ – Z'K May 4 '14 at 9:41
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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: $(x+3)^2 + (y+5)^2 = 25$

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  • $\begingroup$ How did you come up with the solution of finding the radius? I swear to God our professor didn't teach this stuffs that's why I don't have the knowledge with this one. :/ $\endgroup$ – Z'K May 4 '14 at 9:15

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