Answer to Question #276120 in Analytic Geometry for Nick

Question

what is the equation of a circle passing through (6,2) and tangent to the line x-4y-15=0 at (3,-3)

Answer 1

The equation of the circle with center $(a, b)$ and radius $r$ is

$(x-a)^2+(y-b)^2=r^2$

Differentiate both sides with respect to $x$

$2(x-a)+2(y-b)y'=0$

Solve for $y'$

$y'=-\dfrac{x-a}{y-b}$

The slope of the line tangent to the circle at $(3, -3)$

$slope=m=-\dfrac{3-a}{-3-b}=\dfrac{3-a}{3+b}$

The equation of the tangent line tangent to the circle at $(3, -3)$ is

$y-(-3)=\dfrac{3-a}{3+b}(x-3)$ $y=\dfrac{3-a}{3+b}x-3\cdot\dfrac{3-a}{3+b}-3$ $y=\dfrac{3-a}{3+b}x-3\cdot\dfrac{6+b-a}{3+b}$

The line $x-4y-15=0$

$y=\dfrac{1}{4}x-\dfrac{15}{4}$

Then

$\begin{cases} \dfrac{3-a}{3+b}=\dfrac{1}{4}\\ \\ -3\cdot\dfrac{6+b-a}{3+b}=-\dfrac{15}{4} \end{cases}$ $\begin{cases} b=9-4a\\ \\ \dfrac{6+b-a}{3-a}=5 \end{cases}$ $b=9-4a, a\in \R$

Substitute

$(x-a)^2+(y-9+4a)^2=r^2$

Point $(3,-3)$

$(3-a)^2+(-3-9+4a)^2=r^2$ $9-6a+a^2+144-96a+16a^2=r^2$ $r^2=17a^2-102a+153$

Point $(6,2)$

$(6-a)^2+(2-9+4a)^2=r^2$ $36-12a+a^2+49-56a+16a^2=r^2$ $r^2=17a^2-68a+85$ $r^2=17a^2-102a+153=17a^2-68a+85$ $34a=68$ $a=2$ $b=9-4(2)=1$ $r^2=17(2)^2-68(2)+85=17$

The equation of the circle is

$(x-2)^2+(y-1)^2=17$