How to find the equation of a tangent line - SSAT Upper Level Quantitative
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Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
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The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by 
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Find the equation of a tangent line at point 
if the function is 
.
Find the equation of a tangent line at point
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To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function 
is already in the slope-intercept form, 
, and 
.
Substitute the slope and the given point 
into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Tap to see back →
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Find the equation of a tangent line at point if the function is .
Find the equation of a tangent line at point
Tap to see back →
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function is already in the slope-intercept form, , and .
Substitute the slope and the given point into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Tap to see back →
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Find the equation of a tangent line at point if the function is .
Find the equation of a tangent line at point
Tap to see back →
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function is already in the slope-intercept form, , and .
Substitute the slope and the given point into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Tap to see back →
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Find the equation of a tangent line at point if the function is .
Find the equation of a tangent line at point
Tap to see back →
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function is already in the slope-intercept form, , and .
Substitute the slope and the given point into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.