Question 297085: The line drawn tangent to the circle x squared + y squared=169 at (12,5) meets the y-axis where?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Equation of the circle is x^2 + y^2 = 169
The point (12,5) is on the circle.
Since the general equation for a circle is (x-h)^2 + (y-k)^2 = r^2, and (h,k) is the center of the circle, this means that:
The center of the circle is at (h,k) = (0,0)
The radius of the circle is sqrt(169) = 13
The line tangent to the circle at the point (12,5) is perpendicular to the radius of the circle at that point.
the equation for the line that is identical to the radius of the circle (the radius of the circle is a segment of this line bounded by the circumference of the circle) would be given by the equation:
y = m*x + b where m is the slope and b is the y-intercept.
The slope of this line is given by the equation:
(y2-y1) / (x2-x1).
y2 = 5 and y1 = 0
x2 = 12 and x1 = 0
slope of the line = m = (5-0)/(12-0) = 5/12
The equation of this line becomes y = (5/12)*x + b
The y-intercept of this line is found by replacing x and y in the equation by one of the points of the equation and solving for b.
We'll use (x2,y1) = (12,5)
We get:
y = (5/12)x + b becomes 5 = (5/12)*12 + b
Simplify to get:
5 = 5 + b
Subtract 5 from both sides of this equation to get:
b = 0
the y-intercept for this equation is 0 and the equation is:
y = (5/12)*x
This line is identical to the radius of the circle which means that the radius of the circle is part of this line.
The line itself extends indefinitely in both directions.
the radius is a line segment bounded by the circumference of the circle.
The equation of the line perpendicular to that radius will have a slope that is equal to the negative reciprocal of the slope of that line.
This means that the line that represents the tangent to the circle will have a slope of (-12/5)
Since that line passes through the point (12,5), then you can find the y-intercept of that line by replacing y and x in the equation for that line with 5 and 12 to get:
y = (-5/12)x + b becomes 5 = (-12/5)*12 + b
Simplify to get:
5 = -144/5 + b
Multiply both sides of this equation by 5 to get:
25 = -144 + 5b
Add 144 to both sides of this equation to get:
25 + 144 = 5b
Combine like terms to get:
169 = 5b
Divide both sides of this equation by 5 to get:
b = 169/5
the equation for the tangent to the circle passing through the point (12,5) would be:
y = (-12/5)*x + 169/5
169/5 is roughly 33.8 for graphing purposes.
To graph the equation of the circle itself, you have to solve for y.
The equation of the circle is x^2 + y^2 = 169
Solve for y to get y = +/- sqrt(169-x^2)
To graph the circle, you would graph 2 equations.
They would be:
y = sqrt(169-x^2) and y = -sqrt(169-x^2).
Since the equations of the lines are already in slope-intercept form which is the same for required for graphing them, then no change needs to be made to those equations.
You will therefore graph the following equations:
y = (5/12)*x
y = (-12/5)*x + (169/5)
y = sqrt(169-x^2)
y = -sqrt(169-x^2)
The graph of these equations is shown below:
You can see that the line identical to the radius of the circle passes through the points (0,0) and (12,5) and (-12,-5).
The radius is bounded by the points (12,5) and (-12,-5). The line itself extends indefinitely in both directions.
You can see that the line tangent to the circle at the point (12,5) is perpendicular to the radius of the circle at that point and passes through the points (12,5) and (0,33.8). The y-intercept of that line is the point (0,33.8).
A closer look at the tangent point is shown below:
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