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
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To find an equation for the line tangent to the circle at point P, we need to find the derivative of the equation of the circle at point P. The derivative will give us the slope of the tangent line.
The equation of the circle is x^2 + y^2 = 25.
To find the derivative of this equation, we differentiate each term separately with respect to x:
d(x^2)/dx + d(y^2)/dx = d(25)/dx
2x + 2y(dy/dx) = 0
Now we can substitute the values of point P, which is (4, 3), into this equation:
2(4) + 2(3)(dy/dx) = 0
8 + 6(dy/dx) = 0
6(dy/dx) = -8
(dy/dx) = -8/6
(dy/dx) = -4/3
So, the slope of the tangent line at point P is -4/3.
Now, we have the slope of the tangent line and the coordinates of point P, and we need to find the equation of the line. We can use the point-slope form of a line to get the equation:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line, and m is the slope.
Substituting the values of point P and the slope, we get:
y - 3 = (-4/3)(x - 4)
y - 3 = (-4/3)x + 16/3
y = (-4/3)x + 25/3
This is the equation for the line tangent to the circle at point P.
To find the point of tangency on the x-axis, we need to find the x-coordinate when y = 0.
Substituting y = 0 in the equation:
0 = (-4/3)x + 25/3
(4/3)x = 25/3
x = (25/3) * (3/4)
x = 25/4
So, the line meets the x-axis at point Q, which has coordinates (25/4, 0).
Finally, to find the other line through point Q that is tangent to the circle, the process is similar. We can find the slope of a line passing through point Q and use it to determine the equation.
To find the slope, we can use the fact that the radius of a circle is perpendicular to the tangent line. The radius from the center of the circle to point Q is perpendicular to the tangent line at Q, so its slope is the negative reciprocal of the slope of the tangent line.
The center of the circle is at the origin (0, 0), and the coordinates of point Q are (25/4, 0). The slope of the radius is therefore:
m_radius = (0 - 0) / (25/4 - 0)
= 0 / (25/4)
= 0
Since the radius is perpendicular to the tangent line, the slope of the tangent line will be the negative reciprocal of zero, which is undefined.
So, the equation for the other line through point Q that is tangent to the circle is x = 25/4. This line is vertical and intersects the x-axis at point Q, which is also the point of tangency.