Question

The line l is tangent to the circle with equation x^2 + y^2=10 at the point P.

Answer 1

Given:

The equation of a circle is

$x^2+y^2=10$

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

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Endpoints of the radius are O(0,0) and P(1,3). So, the slope of radius is

$m_1=\dfrac{3-0}{1-0}$

$m_1=\dfrac{3}{1}$

$m=3$

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.

$m\times m_1=-1$

$m\times 3=-1$

$m=-\dfrac{1}{3}$

The slope of line l is $-\dfrac{1}{3}$ and it passs through the point P(1,3). So, the equation of line l is

$y-y_1=m(x-x_1)$

$y-3=-\dfrac{1}{3}(x-1)$

$y-3=-\dfrac{1}{3}(x)+\dfrac{1}{3}$

Adding 3 on both sides, we get

$y=-\dfrac{1}{3}x+\dfrac{1}{3}+3$

$y=-\dfrac{1}{3}x+\dfrac{1+9}{3}$

$y=-\dfrac{1}{3}x+\dfrac{10}{3}$

Therefore, the equation of line l is $y=-\dfrac{1}{3}x+\dfrac{10}{3}$.