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(x3)
Question
2 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(x3) 
2. User Answers Blacklash
Answer:
The equation of the line tangent to the circle at point (3,4) is [tex](y4)=\frac{3}{4}(x3)[/tex]
Stepbystep explanation:
First let us find equation of line passing through center of circle (0,0) and point(3,4)
We have
[tex](yy_1)=\frac{y_2y_1}{x_2x_1}(xx_1)\\\\(y0)=\frac{40}{30}(x0)\\\\y=\frac{4}{3}x[/tex]
Slope [tex]=\frac{4}{3}[/tex]
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 [tex]=\frac{1}{\frac{4}{3}}=\frac{3}{4}[/tex]
So the equation of the line tangent to the circle at point (3,4) is given by
[tex](yy_1)=m(xx_1)\\\\(y4)=\frac{3}{4}(x3)[/tex]
The equation of the line tangent to the circle at point (3,4) is [tex](y4)=\frac{3}{4}(x3)[/tex]