Equation of the Tangent to a Circle - Examples with Answers

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 $latex y=-2x+4$.

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 $latex y=-3x+15$.

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 $latex y=-\frac{3}{4}x+\frac{15}{2}$.

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 $latex y=x$.

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 $latex y=-\frac{3}{5}x+\frac{13}{5}$.

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 $latex y=2x+3$.