Answer to Question #234696 in Analytic Geometry for Johnnie

Question

Find the center and radius of the circle that passes through M(3,3) and is tangent to the line Y-2x+3 at point N(5,5)

Answer 1

The line $y=2x+3$ does not pass through the point $N(5,5)$

$$2(5)+3=13\not=5$$

Assume that the problem is:

Find the center and radius of the circle that passes through $M(3,3)$ and is tangent to the line $y=2x+3$ at point $N(1,5)$

The general equation for a circle is $( x - h )^2 + ( y - k )^2 = r^2,$ where $C( h, k)$ is the center and $r$ is the radius.

The circle that passes through $M(3,3)$

$$( 3 - h )^2 + (3 - k )^2 = r^2$$

The circle that passes through $N(1,5)$

$$( 1 - h )^2 + (5 - k )^2 = r^2$$

The line segment $CN$ is perpendicular to the line $y=2x+3$

$$slope_1(slope_2)=-1$$

$$\dfrac{y_N-y_C}{x_N-x_C}(2)=-1$$

$$\dfrac{5-k}{1-h}=-\dfrac{1}{2}$$

$$10-2k=-1+h$$

$$h=11-2k$$

$$( 3 - h )^2 + (3 - k )^2 =( 1 - h )^2 + (5 - k )^2$$

$$9-6h+h^2+9-6k+k^2$$ $$=1-2h+h^2+25-10k+k^2$$

$$18-6h-6k=26-2h-10k$$

$$4h=-8+4k$$

$$h=-2+k$$

Substitute

$$-2+k=11-2k$$

$$k=\dfrac{13}{3}$$

$$h=\dfrac{7}{3}$$

$$( 3 - \dfrac{7}{3} )^2 + (3 - \dfrac{13}{3} )^2 = r^2$$

$$r^2=\dfrac{20}{9}$$

The center of the circle is $C( \dfrac{7}{3},\dfrac{13}{3}).$ The radius of the circle is $\dfrac{2\sqrt{5}}{3}.$

The general equation of the circle is 

$$(x-\dfrac{7}{3})^2 + ( y-\dfrac{13}{3} )^2 = \dfrac{20}{9}$$