The point P = (4, 3) lies on the circle x2 + y2 = 25. Find an equation for the line that is tangent to the circle at P. This line meets the x-axis at a point Q. Find an equation for the other line through Q that is tangent to the circle, and identify its point of tangency
Why did the circle go to the dentist? Because it needed a tangent! Now, let's find that tangent line, shall we?
The equation for the circle is x^2 + y^2 = 25, and the point P lies on it at coordinates (4, 3). To find the equation of the line tangent to the circle at point P, we need to determine its slope.
Differentiating the equation of the circle implicitly with respect to x, we get 2x + 2yy' = 0, where y' represents the derivative of y with respect to x.
Plugging in the coordinates of point P, 4 for x and 3 for y, we have 2(4) + 2(3)y' = 0.
Simplifying further, we get 8 + 6y' = 0, which implies y' = -8/6 or -4/3.
Since the slope of the line tangent to a circle at a given point is equal to the negative reciprocal of the slope of the radius, we can determine the slope of the radius passing through point P. The radius passes through the center of the circle (0, 0) and point P (4, 3), so its slope is (3 - 0) / (4 - 0) = 3/4.
Since the tangent line and radius are perpendicular, and the slope of the tangent is -4/3, we can use these slopes to find the equation of the tangent line.
Using the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we substitute in the values: y - 3 = -4/3(x - 4).
Simplifying the equation, we get: y = -4/3x + 16/3.
Now, let's find the point Q where the tangent line meets the x-axis. To do this, we set y to zero in the equation and solve for x:
0 = -4/3x + 16/3
4/3x = 16/3
x = 4.
Therefore, the point Q where the tangent line meets the x-axis is Q(4, 0).
Now, let's find the equation of the other line through point Q that is also tangent to the circle and identify its point of tangency. Since this line is also tangent to the circle, its slope will also be perpendicular to the radius.
The radius has slope 3/4, which means the slope of the tangent line passing through Q will be -4/3.
Using the point-slope form again, with Q(4, 0) as the point and -4/3 as the slope, we have: y - 0 = -4/3(x - 4).
Simplifying this equation, we get: y = -4/3x + 16/3.
So, the equation of the other line through Q that is tangent to the circle is y = -4/3x + 16/3, and the point of tangency between this line and the circle is P(4, 3).
To find an equation for the line that is tangent to the circle at point P, we can use the following steps:
Step 1: Determine the slope of the line tangent to the circle at point P.
Step 2: Use the slope-intercept form to find the equation of the tangent line.
Step 3: Find the point of intersection of the tangent line with the x-axis to identify point Q.
Step 4: Determine the equation for the other line through Q that is tangent to the circle.
Step 5: Identify the point of tangency of the second tangent line with the circle.
Let's go through each step one by one.
Step 1: Determine the slope of the line tangent to the circle at point P.
The general equation of a circle is x^2 + y^2 = r^2, where (h, k) is the center of the circle, and r is the radius. In this case, the center is at (0, 0), and the radius is √(25) = 5.
To find the slope of the tangent line at point P, we need the slope of the radius connecting the circle's center to point P. The slope of a radius is given by (y2 - y1) / (x2 - x1).
Let's substitute the values:
Slope = (3 - 0) / (4 - 0) = 3/4
Step 2: Use the slope-intercept form to find the equation of the tangent line.
The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept.
Substituting the values we found:
y = (3/4)x + b
Step 3: Find the point of intersection of the tangent line with the x-axis to identify point Q.
To find the point of intersection with the x-axis, we need to set y = 0 in the equation of the tangent line.
0 = (3/4)x + b
Solving for x:
x = -4b/3
This gives us the x-coordinate of point Q.
Step 4: Determine the equation for the other line through Q that is tangent to the circle.
If two lines are tangent to a circle and intersect at the same point, those lines are perpendicular to each other. So the second line that is tangent to the circle will be perpendicular to the tangent line we found in step 2.
The slope of the line perpendicular to the tangent line is the negative reciprocal of the tangent line's slope. So the slope of the second tangent line is -4/3.
Using the point-slope form of a line, where the point of tangency is (x, y) = (4, 3):
y - 3 = (-4/3)(x - 4)
Simplifying the equation:
3y - 9 = -4x + 16
4x + 3y = 25
Step 5: Identify the point of tangency of the second tangent line with the circle.
To find the point of tangency, we can substitute the equation of the second tangent line into the equation of the circle and solve for x and y.
Substituting the equation of the second tangent line into the equation of the circle:
x^2 + y^2 = 25
x^2 + (25 - 4x) = 25
x^2 - 4x + 9 = 0
This equation has no real solutions, so the second tangent line does not intersect the circle.
Therefore, the equation of the line tangent to the circle at point P is y = (3/4)x - 3/4, and the point of tangency is (4, 3).
To find the equation for the line that is tangent to the circle at point P, we can use the concept that the tangent line to a circle is perpendicular to the radius of the circle at the point of tangency.
1. Find the slope of the radius of the circle that passes through point P:
- The center of the circle is at the origin (0, 0), so the radius from the center to point P is given by the coordinates (4, 3).
- The slope of this radius can be found using the formula: slope = (change in y) / (change in x).
slope = (3 - 0) / (4 - 0) = 3/4
2. Since the tangent line is perpendicular to the radius, its slope will be the negative reciprocal of the radius's slope. Therefore, the slope of the tangent line is -4/3.
3. Now that we have the slope of the tangent line and a point on it (P = (4, 3)), we can use the point-slope form of a linear equation to find the equation of the tangent line:
- Point-slope form: y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line, and m is its slope.
- Substituting the values, we get: y - 3 = (-4/3)(x - 4)
- Simplifying the equation: y - 3 = (-4/3)x + 16/3
- Rearranging the equation: y = (-4/3)x + 25/3
4. To find the point where the line meets the x-axis (point Q), we set y = 0 in the equation and solve for x:
0 = (-4/3)x + 25/3
(4/3)x = 25/3
x = 25/4
5. The x-coordinate of point Q is 25/4, and since it lies on the x-axis, the y-coordinate is 0. Therefore, Q = (25/4, 0).
6. Now, to find the equation for the other line through Q that is tangent to the circle, we can use a similar process.
- We already have the slope of the tangent line, which is -4/3.
- We need to find the point of tangency on the circle, which is the intersection of the line through Q and the circle.
- Substituting the coordinates of Q into the equation of the circle, we get: (25/4)² + y² = 25.
Simplifying the equation: 625/16 + y² = 25
y² = 25 - 625/16 = 625/16 - 625/16 = 0
Therefore, y = 0.
7. The x-coordinate of the point of tangency is the same as Q, which is 25/4, and the y-coordinate is 0. Therefore, the point of tangency is (25/4, 0).
8. With the slope of the line (-4/3) and the point of tangency (25/4, 0), we can use the point-slope form again to find the equation of the second tangent line:
y - 0 = (-4/3)(x - 25/4)
Simplifying the equation: y = (-4/3)x + 25/3
So, the equation for the line that is tangent to the circle at P is y = (-4/3)x + 25/3, and the equation for the other line through Q that is tangent to the circle is y = (-4/3)x + 25/3. The point of tangency for both lines is (25/4, 0).
If the center is at O, then the slope of OP = 3/4
So, the tangent line, which is perpendicular to the radius OP has slope -4/3, and its equation is thus
y-3 = -4/3 (x-4)
The line's x-intercept is at Q=(7,0)
Due to symmetry, the other tangent line through Q touches the circle at (-3,4), so its equation is
y+3 = 4/3 (x-4)
see
http://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%3D25,+y+%3D+-4%2F3+(x-4)%2B3,+y+%3D+4%2F3+(x-4)-3,+y%3D0