# 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)

Mathematics

## 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?

• ### 1. User Answers pepe11

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)
• ### 2. User Answers Blacklash

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