Tangent Lines - Intermediate Geometry

Card 0 of 12

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

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

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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:

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

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

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

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

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

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

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

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Question

What is the slope of the tangent line to the graph of when ?

Answer

To find the slope of the tangent line, first we must take the derivative of , giving us . Next we simply plug in our given x-value, which in this case is . This leaves us with a slope of .

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Question

Suppose the equation of a line is . What is the slope of the tangent line at ?

Answer

Rewrite in slope-intercept form, .

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

The slope of the tangent line is $\frac{3}{5}$ .

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Question

Suppose a function . What is the slope of the tangent line at ?

Answer

Write the formula for slope-intercept form.

The slope of is always zero at every point on the domain. Therefore, the slope at must also be zero.

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