Question

Consider the circle of radius 5 centered at (0,0). Find an equation of the line tangent to the circle at point (3,4). The book gives me the answer: y=-3/4(x-3) but can you explain to me how you get this answer?

Answer 1

Answer:

The equation of the line tangent to the circle at point (3,4) is $(y-4)=-\frac{3}{4}(x-3)$

Step-by-step explanation:

First let us find equation of line passing through center of circle (0,0) and point(3,4)

We have

             $(y-y_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(y-0)=\frac{4-0}{3-0}(x-0)\\\\y=\frac{4}{3}x$

         Slope $=\frac{4}{3}$

This line is perpendicular to line tangent to the circle at point (3,4).

Product of slopes of perpendicular lines = -1

Slope of tangent line $=\frac{-1}{\frac{4}{3}}=-\frac{3}{4}$

So the equation of the line tangent to the circle at point (3,4) is given by

             $(y-y_1)=m(x-x_1)\\\\(y-4)=-\frac{3}{4}(x-3)$

The equation of the line tangent to the circle at point (3,4) is $(y-4)=-\frac{3}{4}(x-3)$

         

Answer 2

Equation of the circle: x² + y² = 25
9 + 16 = 25 => the point (3,5) is on the circle.
the line tangent to the circle is perpendicular to the radius O point
slope of the radius: 4/3 => slope of the tangent = -3/4
tangent contain (3,4) and have -3/4 as slope.
so equation is : y - 4 = -3/4 (x - 3) => y = -3/4x - 4 + 9/4 + 4 => y = -3/4x + 9/4 
or y = -3/4(x-3)