How to find the equation of a tangent line - Geometry

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Question

A line is tangent to the circle $(x-4)^{2} + (y-2)^{2} = 100$ at the point

What is the equation of this line?

Answer

The center of this circle is

Therefore, the radius with endpoint has slope

$\frac{2-(-6)}{4-10}= \frac{8}{-6}= -\frac{4}{3}$

The tangent line at is perpendicular to this radius; therefore, its slope is the opposite of the reciprocal of $-\frac{4}{3}$ , or $\frac{3}{4}$ .

Now use the point-slope formula with this slope and the point of tangency:

$y-(-6)= \frac{3}{4} \left ( x-10\right )$

$y+6= \frac{3}{4}x-\frac{15}{2}$

$y= \frac{3}{4}x-\frac{27}{2}$

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Question

Given the function , find the equation of the tangent line passing through .

Answer

Find the slope of . The slope is 3.

Substitute to determine the y-value.

The point is .

Use the slope-intercept formula to find the y-intercept, given the point and slope.

Substitute the point and the slope.

Substitute the y-intercept and slope back to the slope-intercept formula.

The correct answer is:

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Question

Find the equation of the tangent line at the point if the given function is .

Answer

Write in slope-intercept form and determine the slope.

Rearranging our equation to be in slope-intercept form we get:

$y=\frac{3}{2}x+2$ .

The slope is our value which is $\frac{3}{2}$ .

Substitute the slope and the point to the slope-intercept form.

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

Substitute the slope and y-intercept to find our final equation.

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

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Question

What is the equation of the tangent line at to the equation ?

Answer

Rewrite in slope-intercept form to determine the slope. Remember slope-intercept form is .

Therefore, the equation becomes,

$y=\frac{3}{5}x-\frac{2}{5}$ .

The slope is the value of the function thus, it is $\frac{3}{5}$ .

Substitute back to the original equation to find the value of .

$y=-\frac{2}{5}$

Substitute the point $\left(0,-\frac{2}{5}\right)$ and the slope of the line into the slope-intercept equation.

$-\frac{2}{5}=\left(\frac{3}{5}\right)(0)+b$

$b=-\frac{2}{5}$

Substitute the point and the slope back in to the slope-intercept formula.

$y=\frac{3}{5}x-\frac{2}{5}$

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Question

Write the equation for the tangent line of the circle through the point .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 3 - - 2 }{ 0 - 4 } = \frac{ 5 }{ -4 }$

Since the tangent line is perpendicular, its slope is $\frac{4}{5 }$

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

$-2 = \frac{4}{5}(4) + b$

$-2 = \frac{16} { 5 } + b$

$-\frac{10}{5} = \frac{16} {5} + b$

$-\frac{26}{5} = b$

The equation is $y = \frac{4}{5} x - \frac{26} {5 }$

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Question

Find the equation for the tangent line at for the circle .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 8-2 }{ -3+4 } = \frac{ 6 }{ 1 }$

Since the tangent line is perpendicular, its slope is $-\frac{1}{6 }$

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

$8= -\frac{1}{6}(-3) + b$

$8 = \frac{1} { 2 } + b$

$7 \frac{1}{2} = b$

The equation is $y = -\frac{1}{6} x+7 \frac{1}{2}$

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Question

Find the equation for the tangent line to the circle at the point .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 3 --2 }{ 15 -3 } = \frac{ 5 }{ 12 }$

Since the tangent line is perpendicular, its slope is $-\frac{12}{5 }$

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

$3 = -\frac{12}{5}(15) + b$

The equation is $y = - \frac{12}{5} x+39$

Compare your answer with the correct one above

Question

Find the equation for the tangent line of the circle at the point .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{-1 - 3 }{ 4 - - 2 } = \frac{-4}{6 } = -\frac{2}{3}$

Since the tangent line is perpendicular, its slope is $\frac{3}{2 }$ .

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

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

The equation is $y = \frac{3}{2} x +6$

Compare your answer with the correct one above

Question

Refer to the above diagram,

Give the equation of the line tangent to the circle at the point shown.

Answer

The tangent to a circle at a given point is perpendicular to the radius that has the center and the given point as its endpoints.

The circle has its center at origin ; since this and are the endpoints of the radius - and the line that includes this radius includes both points - its slope can be found by setting $x_{1} = y_{1} = 0 , x_{2}= 4, y_{2} = 9$ in the following slope formula:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m = \frac{9-0}{4-0} = \frac{9 }{4 }$

The tangent line, being perpendicular to this radius, has as its slope the opposite of the reciprocal of this, which is $m = -\frac{4}{9}$ . Since the tangent line includes point , set $m = -\frac{4}{9}, x_{1} = 4, y_{1} = 9$ in the point-slope formula and simplify:

$y - y_{1} = m (x-x_{1})$

$y - 9= -\frac{4}{9} (x-4)$

$y - 9= -\frac{4}{9} \cdot x- \left (-\frac{4}{9} \right ) \cdot 4$

$y - 9= -\frac{4}{9} x+ \frac{16}{9}$

$y - 9+ 9 = -\frac{4}{9} x+ \frac{16}{9} + 9$

$y = -\frac{4}{9} x+ \frac{16}{9} + \frac{81}{9}$

$y = -\frac{4}{9} x+ \frac{97}{9}$

Compare your answer with the correct one above

Question

A line is tangent to the circle $(x-4)^{2} + (y-2)^{2} = 100$ at the point

What is the equation of this line?

Answer

The center of this circle is

Therefore, the radius with endpoint has slope

$\frac{2-(-6)}{4-10}= \frac{8}{-6}= -\frac{4}{3}$

The tangent line at is perpendicular to this radius; therefore, its slope is the opposite of the reciprocal of $-\frac{4}{3}$ , or $\frac{3}{4}$ .

Now use the point-slope formula with this slope and the point of tangency:

$y-(-6)= \frac{3}{4} \left ( x-10\right )$

$y+6= \frac{3}{4}x-\frac{15}{2}$

$y= \frac{3}{4}x-\frac{27}{2}$

Compare your answer with the correct one above

Question

Given the function , find the equation of the tangent line passing through .

Answer

Find the slope of . The slope is 3.

Substitute to determine the y-value.

The point is .

Use the slope-intercept formula to find the y-intercept, given the point and slope.

Substitute the point and the slope.

Substitute the y-intercept and slope back to the slope-intercept formula.

The correct answer is:

Compare your answer with the correct one above

Question

Find the equation of the tangent line at the point if the given function is .

Answer

Write in slope-intercept form and determine the slope.

Rearranging our equation to be in slope-intercept form we get:

$y=\frac{3}{2}x+2$ .

The slope is our value which is $\frac{3}{2}$ .

Substitute the slope and the point to the slope-intercept form.

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

Substitute the slope and y-intercept to find our final equation.

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

Compare your answer with the correct one above

Question

What is the equation of the tangent line at to the equation ?

Answer

Rewrite in slope-intercept form to determine the slope. Remember slope-intercept form is .

Therefore, the equation becomes,

$y=\frac{3}{5}x-\frac{2}{5}$ .

The slope is the value of the function thus, it is $\frac{3}{5}$ .

Substitute back to the original equation to find the value of .

$y=-\frac{2}{5}$

Substitute the point $\left(0,-\frac{2}{5}\right)$ and the slope of the line into the slope-intercept equation.

$-\frac{2}{5}=\left(\frac{3}{5}\right)(0)+b$

$b=-\frac{2}{5}$

Substitute the point and the slope back in to the slope-intercept formula.

$y=\frac{3}{5}x-\frac{2}{5}$

Compare your answer with the correct one above

Question

Write the equation for the tangent line of the circle through the point .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 3 - - 2 }{ 0 - 4 } = \frac{ 5 }{ -4 }$

Since the tangent line is perpendicular, its slope is $\frac{4}{5 }$

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

$-2 = \frac{4}{5}(4) + b$

$-2 = \frac{16} { 5 } + b$

$-\frac{10}{5} = \frac{16} {5} + b$

$-\frac{26}{5} = b$

The equation is $y = \frac{4}{5} x - \frac{26} {5 }$

Compare your answer with the correct one above

Question

Find the equation for the tangent line at for the circle .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 8-2 }{ -3+4 } = \frac{ 6 }{ 1 }$

Since the tangent line is perpendicular, its slope is $-\frac{1}{6 }$

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

$8= -\frac{1}{6}(-3) + b$

$8 = \frac{1} { 2 } + b$

$7 \frac{1}{2} = b$

The equation is $y = -\frac{1}{6} x+7 \frac{1}{2}$

Compare your answer with the correct one above

Question

Find the equation for the tangent line to the circle at the point .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 3 --2 }{ 15 -3 } = \frac{ 5 }{ 12 }$

Since the tangent line is perpendicular, its slope is $-\frac{12}{5 }$

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

$3 = -\frac{12}{5}(15) + b$

The equation is $y = - \frac{12}{5} x+39$

Compare your answer with the correct one above

Question

Find the equation for the tangent line of the circle at the point .

Answer

The circle's center is . The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{-1 - 3 }{ 4 - - 2 } = \frac{-4}{6 } = -\frac{2}{3}$

Since the tangent line is perpendicular, its slope is $\frac{3}{2 }$ .

To write the equation in the form , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point for x and y:

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

The equation is $y = \frac{3}{2} x +6$

Compare your answer with the correct one above

Question

Refer to the above diagram,

Give the equation of the line tangent to the circle at the point shown.

Answer

The tangent to a circle at a given point is perpendicular to the radius that has the center and the given point as its endpoints.

The circle has its center at origin ; since this and are the endpoints of the radius - and the line that includes this radius includes both points - its slope can be found by setting $x_{1} = y_{1} = 0 , x_{2}= 4, y_{2} = 9$ in the following slope formula:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m = \frac{9-0}{4-0} = \frac{9 }{4 }$

The tangent line, being perpendicular to this radius, has as its slope the opposite of the reciprocal of this, which is $m = -\frac{4}{9}$ . Since the tangent line includes point , set $m = -\frac{4}{9}, x_{1} = 4, y_{1} = 9$ in the point-slope formula and simplify:

$y - y_{1} = m (x-x_{1})$

$y - 9= -\frac{4}{9} (x-4)$

$y - 9= -\frac{4}{9} \cdot x- \left (-\frac{4}{9} \right ) \cdot 4$

$y - 9= -\frac{4}{9} x+ \frac{16}{9}$

$y - 9+ 9 = -\frac{4}{9} x+ \frac{16}{9} + 9$

$y = -\frac{4}{9} x+ \frac{16}{9} + \frac{81}{9}$

$y = -\frac{4}{9} x+ \frac{97}{9}$

Compare your answer with the correct one above

Question

A line is tangent to the circle $(x-4)^{2} + (y-2)^{2} = 100$ at the point

What is the equation of this line?

Answer

The center of this circle is

Therefore, the radius with endpoint has slope

$\frac{2-(-6)}{4-10}= \frac{8}{-6}= -\frac{4}{3}$

The tangent line at is perpendicular to this radius; therefore, its slope is the opposite of the reciprocal of $-\frac{4}{3}$ , or $\frac{3}{4}$ .

Now use the point-slope formula with this slope and the point of tangency:

$y-(-6)= \frac{3}{4} \left ( x-10\right )$

$y+6= \frac{3}{4}x-\frac{15}{2}$

$y= \frac{3}{4}x-\frac{27}{2}$

Compare your answer with the correct one above

Question

Given the function , find the equation of the tangent line passing through .

Answer

Find the slope of . The slope is 3.

Substitute to determine the y-value.

The point is .

Use the slope-intercept formula to find the y-intercept, given the point and slope.

Substitute the point and the slope.

Substitute the y-intercept and slope back to the slope-intercept formula.

The correct answer is:

Compare your answer with the correct one above

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