# Tangent to a Circle Homework Solution

• sooyong94
In summary: Note that we can't use the letters a and b as constants in a quadratic equation in x since those letters are already being used as the coordinates of the center of the circle.)Now you can use the quadratic formula to solve for x in terms of m, c, b, and r. Since the line y = mx + c is tangent to the circle, there will be exactly one real value of x that satisfies the equation. That is, the discriminant b^2 - 4ac = 0. (If the line were secant to the circle, there would be two real values of x that satisfy the equation, and if the line were not tangent to the circle, there would be no real values of x that
sooyong94

## Homework Statement

Given that the line ##y=mx+c## is a tangent to the circle ##(x-a)^{2} +(y-b)^{2} =r^{2}##, show that ##(1+m^{2}) r^{2}=(c-b+ma)^{2}##

## Homework Equations

Quadratic discriminant, sum and product of roots

## The Attempt at a Solution

I substituted y=mx+c into the equation of the circle, and this is what I have:
##(x-a)^{2} +(mx+c-b)^{2} =r^{2}##
Then I expanded and simplified into this:
##(1+m^{2})x^{2}+(2mc-2mb-2a)x+a^{2}+b^{2}-2bc+c^{2}-r^{2}=0##

Now if I have to use the quadratic discriminant it would be tedious to work with, unfortunately. :(

Last edited by a moderator:
sooyong94 said:

## Homework Statement

Given that the line ##y=mx+c## is a tangent to the circle ##(x-a)^{2} +(y-b)^{2} =r^{2}##, show that ##(1+m^{2}) r^{2}=(c-b+ma)^{2}

## Homework Equations

Quadratic discriminant, sum and product of roots

## The Attempt at a Solution

I substituted y=mx+c into the equation of the circle, and this is what I have:
##(x-a)^{2} +(mx+c-b)^{2} =r^{2}##
Then I expanded and simplified into this:
##(1+m^{2})x^{2}+(2mc-2mb-2a)x+a^{2}+b^{2}-2bc+c^{2}-r^{2}=0##

Now if I have to use the quadratic discriminant it would be tedious to work with, unfortunately. :(

There's a lot of Algebraic manipulation in this problem.

First hint: Find m, which is the same as ##\frac{dy}{dx}##
Second hint: solve for c in terms of only a, b, x, y
Third hint: Show both sides of the equations are equal to each other (again, only in terms of a, b, x, y).

Edit: This is in the Precalculus section. Have you learned how to find ##\frac{dy}{dx}## yet?

Last edited:
Use the concept that the radius of circle is equal to the distance of the given line from the centre of circle. You already know the formula for finding the distance of any point from a line and in this case it is simply the centre of circle (a,b). Equate both and you're done.

sooyong94 said:

## Homework Statement

Given that the line ##y=mx+c## is a tangent to the circle ##(x-a)^{2} +(y-b)^{2} =r^{2}##, show that ##(1+m^{2}) r^{2}=(c-b+ma)^{2}

## Homework Equations

Quadratic discriminant, sum and product of roots

## The Attempt at a Solution

I substituted y=mx+c into the equation of the circle, and this is what I have:
##(x-a)^{2} +(mx+c-b)^{2} =r^{2}##
Then I expanded and simplified into this:
##(1+m^{2})x^{2}+(2mc-2mb-2a)x+a^{2}+b^{2}-2bc+c^{2}-r^{2}=0##

Now if I have to use the quadratic discriminant it would be tedious to work with, unfortunately. :(

You should have expanded only so far as
$$(1 + m^2)x^2 + 2(m(c-b) - a)x + a^2 + (c-b)^2 - r^2 = 0$$
Keeping (c-b) as (c-b) throughout will make your calculations much less tedious.

## What is a tangent to a circle?

A tangent to a circle is a line that touches the circle at exactly one point, called the point of tangency. It is perpendicular to the radius of the circle at that point.

## How do you find the equation of a tangent to a circle?

The equation of a tangent to a circle can be found by using the formula y = mx + b, where m is the slope of the tangent line and b is the y-intercept. The slope of the tangent line is equal to the negative reciprocal of the slope of the radius at the point of tangency. The y-intercept can be found by substituting the coordinates of the point of tangency into the equation.

## What is the relationship between the radius and the tangent line to a circle?

The radius of a circle and the tangent line to the circle are always perpendicular to each other at the point of tangency. This means that the tangent line forms a right angle with the radius at that point.

## Can a circle have more than one tangent line?

Yes, a circle can have infinitely many tangent lines. Any line that passes through the point of tangency and is perpendicular to the radius can be considered a tangent line to the circle.

## How is the length of a tangent line to a circle calculated?

The length of a tangent line to a circle can be calculated using the Pythagorean theorem. The tangent line, radius, and a line connecting the center of the circle to the point of tangency form a right triangle. Using the Pythagorean theorem, the length of the tangent line can be found by taking the square root of the sum of the squares of the radius and the distance from the center of the circle to the point of tangency.

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