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

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