The equation of the tangent line to a circle is found using the form *y*=*mx*+*b*. In turn, we can find the slope, *m*, by determining the slope of the radius using the center of the circle and the tangential point. Then, we use the tangential point to find the value of *b*.

Below, we will learn how to find the equation of the tangent to a circle step by step. Then, we will apply this process to solve some practice exercises.

## How to find the equation of the tangent line to a circle

To find the equation of the tangent to a circle, we recall that we can find the equation of any line using the form $latex y=mx+b$, where *m* is the slope and *b* is the *y*-intercept.

In this case, the slope *m* is found using the slope of the radius, and the *y*-intercept is found using the coordinates of the tangent point.

Then, we follow the next steps:

#### 1. Find the coordinates of the center of the circle.

If the equation is given in its standard form, $latex r^2=(x-a)^2+(x-b)^2$, the center is $latex (a,~b)$. If we have a general equation of the circle, we have to complete the square of *x* and *y* to find the center.

#### 2. Find the slope of the radius of the circle.

For this, we use the slope formula $latex m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ with the coordinates of the tangential point and the center of the circle.

#### 3. Find the slope of the tangent line.

Since the radius and tangent are perpendicular, the slope of the tangent is equal to the negative reciprocal of the slope of the radius.

#### 4. Find the value of *b*, the *y*-intercept.

For this, we use the slope from step 3 and the coordinates of the tangential point in the form $latex y=mx+b$ and solve for *b*.

## Equation of the tangent to a circle – Examples with answers

**EXAMPLE 1**

Find the equation of the tangent line to the circle $latex (x+2)^2+(y-3)^2=4$ at the point $latex P=(0,~4)$.

##### Solution

**Step 1:** Since the equation of the circle is in standard form, we can easily see that the center is equal to $latex (-2, 3)$.

**Step 2:** Use $latex (x_{1},~y_{1})=(-2, 3)$ and $latex (x_{2},~y_{2})=(0, ~4)$ to find the slope of the radius of the circle:

$$m_{r}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$=\frac{4-3}{0+2}$$

$$=\frac{1}{2}$$

**Step 3:** The slope of the tangent line is:

$$m=-\frac{1}{m_{r}}$$

$$m=-2$$

**Step 4:** To find the value of

*b*, we use $latex (x,~y)=(0,~4)$:

$latex y=mx+b$

$latex 4=-2(0)+b$

$latex b=4$

The equation of the tangent line to the circle at $latex (0, ~4)$ is $latex y=-2x+4$.

**EXAMPLE **2

**EXAMPLE**What is the equation of the tangent line to the circle $latex (x-1)^2+(y-2)^2=10$ at the point $latex P=(4,~3)$?

##### Solution

**Step 1:** We have the equation of the circle in its standard form, so we see that the center is $latex (1, 2)$.

**Step 2:** Find the slope of the radius of the circle using the coordinates $latex (x_{1},~y_{1})=(1, 2)$ and $latex (x_{2},~y_{2})=(4, ~3)$ to find the slope of the radius of the circle:

$$m_{r}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$=\frac{3-2}{4-1}$$

$$=\frac{1}{3}$$

**Step 3:** The slope of the tangent line is:

$$m=-\frac{1}{m_{r}}$$

$$m=-3$$

**Step 4:** Using the tangential point $latex (x,~y)=(4,~3)$, we find the value of

*b*:

$latex y=mx+b$

$latex 3=-3(4)+b$

$latex b=15$

The equation of the tangent line to the circle at $latex (4, ~3)$ is $latex y=-3x+15$.

**EXAMPLE **3

**EXAMPLE**Find the equation of the tangent line to the circle $latex x^2+y^2+2x-4y-20=0$ at the point $latex P=(2, 6)$.

##### Solution

**Step 1:** In this case, the equation of the circle is in its general form. Then, we write it in its standard form by completing the square of both variables:

$latex x^2+y^2+2x-4y-20=0$

$latex x^2+2x+y^2-4y-20=0$

$$(x+1)^2-1+(y-2)^2-4-20=0$$

$latex (x+1)^2+(y-2)^2=25$

Now, we see that the center is $latex (-1, ~2)$.

**Step 2:** The slope of the radius of the circle is:

$$m_{r}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$=\frac{6-2}{2+1}$$

$$=\frac{4}{3}$$

**Step 3:** The slope of the tangent line is:

$$m=-\frac{1}{m_{r}}$$

$$m=-\frac{3}{4}$$

**Step 4:** To find the value of

*b*, we use the coordinates of the tangent point $latex (x,~y)=(2,~6)$:

$latex y=mx+b$

$latex 6=-\frac{3}{4}(2)+b$

$latex b=\frac{15}{2}$

The equation of the tangent line to the circle at $latex (2, ~6)$ is $latex y=-\frac{3}{4}x+\frac{15}{2}$.

**EXAMPLE **4

**EXAMPLE**What is the equation of the tangent line to the circle $latex x^2+y^2-2x-6y+8=0$ at the point $latex P=(2, 2)$?

##### Solution

**Step 1:** Writing the equation of the circle in its standard form, we have:

$latex x^2+y^2-2x-6y+8=0$

$latex x^2-2x+y^2-6y+8=0$

$$(x-1)^2-1+(y-3)^2-9+8=0$$

$latex (x-1)^2+(y-3)^2=2$

The center of the circle is $latex (1, ~3)$.

**Step 2:** Finding the slope of the radius of the circle, we have:

$$m_{r}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$=\frac{2-3}{2-1}$$

$latex =-1$

**Step 3:** The slope of the tangent line is:

$$m=-\frac{1}{m_{r}}$$

$$m=1$$

**Step 4:** Use the tangent point to find the value of

*b*:

$latex y=mx+b$

$latex 2=1(2)+b$

$latex b=0$

The equation of the tangent line to the circle at $latex (2, ~2)$ is $latex y=x$.

**EXAMPLE **5

**EXAMPLE**Find the equation of the tangent line to the circle $latex x^2+y^2+4x+6y-21=0$ at the point $latex P=(1, ~2)$.

##### Solution

**Step 1:** Finding the standard form of the equation of the given circle, we have:

$latex x^2+y^2+4x+6y-21=0$

$latex x^2+4x+y^2+6y-21=0$

$$(x+2)^2-4+(y+3)^2-9-21=0$$

$latex (x+2)^2+(y+3)^2=34$

The center of the circle is $latex (-2,~-3)$.

**Step 2:** Finding the slope of the radius of the circle, we have:

$$m_{r}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$=\frac{2+3}{1+2}$$

$$=\frac{5}{3}$$

**Step 3:** The slope of the tangent line is:

$$m=-\frac{1}{m_{r}}$$

$$m=-\frac{3}{5}$$

**Step 4:** Use$latex (x,~y)=(1,~2)$ to find he value of

*b*:

$latex y=mx+b$

$latex 2=-\frac{3}{5}(1)+b$

$latex b=\frac{13}{5}$

The equation of the tangent line to the circle at $latex (1, ~2)$ is $latex y=-\frac{3}{5}x+\frac{13}{5}$.

**EXAMPLE **6

**EXAMPLE**What is the equation of the tangent line to the circle $latex x^2+y^2+6x-4y+8=0$ at the point $latex P=(-1, 1)$?

##### Solution

**Step 1:** Writing the equation of the circle in its standard form, we have:

$latex x^2+y^2+6x-4y+8=0$

$latex x^2+6x+y^2-4y+8=0$

$$(x+3)^2-9+(y-2)^2-4+8=0$$

$latex (x+3)^2+(y-2)^2=5$

Now, we see that the center of the circle is $latex (-3, ~2)$.

**Step 2:** The slope of the radius of the circle is:

$$m_{r}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$=\frac{1-2}{-1+3}$$

$$=-\frac{1}{2}$$

**Step 3:** The slope of the tangent line is:

$$m=-\frac{1}{m_{r}}$$

$latex m=2$

**Step 4:** To find the value of

*b*, we use $latex (x,~y)=(-1,~1)$:

$latex y=mx+b$

$latex 1=2(-1)+b$

$latex b=3$

The equation of the tangent line to the circle at $latex (-1, ~1)$ is $latex y=2x+3$.

## Equation of the tangent to a circle – Practice problems

#### Find the equation of the tangent to the following circle at the point (1, 4): $$ x^2-6x+y^2-6y+18=5$$

Write the equation in the input box.

## See also

Interested in learning more about circles? You can look at these pages:

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