30.6k views
3 votes
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?

User Kirstie
by
7.7k points

2 Answers

5 votes
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)
User Ezeke
by
7.6k points
6 votes

Answer:

The equation of the line tangent to the circle at point (3,4) is
(y-4)=-(3)/(4)(x-3)

Explanation:

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

We have


(y-y_1)=(y_2-y_1)/(x_2-x_1)(x-x_1)\\\\(y-0)=(4-0)/(3-0)(x-0)\\\\y=(4)/(3)x

Slope
=(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
=(-1)/((4)/(3))=-(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)=-(3)/(4)(x-3)

The equation of the line tangent to the circle at point (3,4) is
(y-4)=-(3)/(4)(x-3)

User DhruvPathak
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories