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. :/
[Math] Finding an equation of a circle with a given center and a tangent line.
analytic geometrycircles
Related Solutions
There is a general expression derived in HCR's Web Papers for directly calculating the co-ordinates of foot of perpendicular say $(x', y')$ drawn from any point $(x_o, y_o)$ to the straight line $y=mx+c$ is given as $$(x', y')\equiv\left(\frac{x_o+m(y_o-c)}{1+m^2}, \frac{mx_o+m^2y_o+c}{1+m^2}\right)$$ Hence, the co-ordinate of the other end of diameter i.e. foot of perpendicular $(x', y')$ drawn from the point $(4, 1)$ to the line y=3x-1 are calculated as follows $$(x', y')\equiv\left(\frac{4+3(1-(-1))}{1+3^2}, \frac{3(4)+3^2(1)+(-1)}{1+3^2}\right)\equiv\left(\frac{10}{10}, \frac{20}{10} \right)\equiv\left(1, 2\right)$$ Hence, the co-ordinates of the center of the circle will be the mid-point of the line joining the end-points of the diameter $(4, 1)$ & $(1, 2)$ are calculated as follows $$\left(\frac{4+1}{2}, \frac{1+2}{2}\right)\equiv\left(\frac{5}{2}, \frac{3}{2}\right)$$
There are obviously not a unique one but multiple ( an infinite set) circle solutions.
They can be set by a variable center point given on the given parametrized straight line
$$ (x_C,y_C)=(t, \frac{5-t}{4})$$
You can find them all
$$ (x-x_C)^2+(y-y_C)^2= y_C^2.$$
Find the equation of Circle in the question misleads.
Best Answer
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 $$(x+3)^2+(x+5)^2=5^2$$