Given:
The equation of a circle is
[tex]x^2+y^2=10[/tex]
A tangent line l to the circle touches the circle at point P(1,3).
To find:
The equation 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(1,3). So, the slope of radius is
[tex]m_1=\dfrac{3-0}{1-0}[/tex]
[tex]m_1=\dfrac{3}{1}[/tex]
[tex]m=3[/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 3=-1[/tex]
[tex]m=-\dfrac{1}{3}[/tex]
The slope of line l is [tex]-\dfrac{1}{3}[/tex] and it passs through the point P(1,3). So, the equation of line l is
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-3=-\dfrac{1}{3}(x-1)[/tex]
[tex]y-3=-\dfrac{1}{3}(x)+\dfrac{1}{3}[/tex]
Adding 3 on both sides, we get
[tex]y=-\dfrac{1}{3}x+\dfrac{1}{3}+3[/tex]
[tex]y=-\dfrac{1}{3}x+\dfrac{1+9}{3}[/tex]
[tex]y=-\dfrac{1}{3}x+\dfrac{10}{3}[/tex]
Therefore, the equation of line l is [tex]y=-\dfrac{1}{3}x+\dfrac{10}{3}[/tex].
