Question

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

Determine the equation of line l.

Answer 1

The equation of the tangent line l to the circle x^2 + y^2 = 10 at point P(1,3) is y = -1/3x + 10/3.

To find the equation of the tangent line l to the circle x^2 + y^2 = 10 at point P(1,3), we follow these steps.

The general equation of a circle is x^2 + y^2 = r^2, where r is the radius. In this case, the circle has a radius of √10 because r^2 = 10.

Now, let's find the slope of the radius from the center of the circle (the origin) to point P. The slope of the radius is given by the change in y divided by the change in x. For the point P(1,3), the slope is (3 - 0)/(1 - 0) = 3.

The negative reciprocal of this slope gives us the slope of the tangent line l. So, the slope of l is -1/3.

Now, we use the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope.

Substitute the values for x1, y1, and m:

y - 3 = -1/3(x - 1)

Now, simplify and rearrange the equation:

y = -1/3x + 10/3

So, the equation of the tangent line l to the circle at point P(1,3) is y = -1/3x + 10/3.

Answer 2

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=(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=(3-0)/(1-0)`

`m_1=(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* m_1=-1`

`m* 3=-1`

`m=-(1)/(3)`

The slope of line l is
`-(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=-(1)/(3)(x-1)`

`y-3=-(1)/(3)(x)+(1)/(3)`

Adding 3 on both sides, we get

`y=-(1)/(3)x+(1)/(3)+3`

`y=-(1)/(3)x+(1+9)/(3)`

`y=-(1)/(3)x+(10)/(3)`

Therefore, the equation of line l is
`y=-(1)/(3)x+(10)/(3)`.