Answer to Question #277298 in Analytic Geometry for Zizu

Question

Find the equation of the line tangent to the circle with the center at (-1,1) and point of tangency at (-1,3).

Answer 1

The equation of the circle

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

The point $(-1,3)$ lies on the circle

$$(-1-(-1))^2+(3-1)^2=R^2$$

$$R^2=4$$

$$(x+1)^2+(y-1)^2=4$$

Differentiate both sides with respect to $x$

$$2(x+1)+2(y-1)y'=0$$

Solve for $y'$

$$y'=-\dfrac{x+1}{y-1}$$

Point of tangency at $(-1,3).$

$$y'=-\dfrac{-1+1}{3-1}=0$$

The equation of the tangent in point-slope form

$$y-3=0(x-(-1))$$

The equation of the tangent in slope-intercept form

$$y=3$$