Given:
The equation of a circle is
[tex]x^2+y^2=169[/tex]
A tangent line l to the circle touches the circle at point P(12,5).
To find:
The gradient of the line l.
Solution:
Slope formula: If a line passes through two points, then the slope of the line is
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Endpoints of the radius are O(0,0) and P(12,5). So, the slope of radius is
[tex]m_1=\dfrac{5-0}{12-0}[/tex]
[tex]m_1=\dfrac{5}{12}[/tex]
We know that, the radius of a circle is always perpendicular to the tangent at the point of tangency.
Product of slopes of two perpendicular lines is always -1.
Let the slope of tangent line l is m. Then, the product of slopes of line l and radius is -1.
[tex]m\times m_1=-1[/tex]
[tex]m\times \dfrac{5}{12}=-1[/tex]
[tex]m=-\dfrac{12}{5}[/tex]
Therefore, the gradient or slope of the tangent line l is [tex]-\dfrac{12}{5}[/tex] .
