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Mariana [72]
2 years ago
10

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

Mathematics
1 answer:
babunello [35]2 years ago
4 0

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}.

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