Question #277298

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

1
Expert's answer
2021-12-09T07:43:05-0500

The equation of the circle


(x(1))2+(y1)2=R2(x-(-1))^2+(y-1)^2=R^2

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


(1(1))2+(31)2=R2(-1-(-1))^2+(3-1)^2=R^2

R2=4R^2=4

(x+1)2+(y1)2=4(x+1)^2+(y-1)^2=4

Differentiate both sides with respect to xx


2(x+1)+2(y1)y=02(x+1)+2(y-1)y'=0

Solve for yy'


y=x+1y1y'=-\dfrac{x+1}{y-1}

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


y=1+131=0y'=-\dfrac{-1+1}{3-1}=0

The equation of the tangent in point-slope form


y3=0(x(1))y-3=0(x-(-1))

The equation of the tangent in slope-intercept form


y=3y=3

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