From an exteror point P that is h units form a circle of radius r, a tangent line is drawn to the circle. Let d denote the distance from the point P to the point of tangency T. Express d as a function of h.(HINT: if C is the center of the circle, the PT is perpendicular to CT)
To find the distance from the point P to the point of tangency T, we can use the Pythagorean theorem.
Step 1: Draw a diagram of the situation described. Label the center of the circle as C, the point of tangency as T, and the distance from P to T as d.
Step 2: Draw a line segment from the center of the circle C to the point of tangency T. This line segment will be perpendicular to the tangent line.
Step 3: Notice that we now have a right triangle formed by sides CT, PT, and PC.
Step 4: According to the Pythagorean theorem, the square of the hypotenuse (CT) is equal to the sum of the squares of the other two sides (PT and PC).
Step 5: Since the line segment PT is the distance we want to find, we can express it as d^2.
Step 6: The length of the side PC is equal to the radius of the circle, which is r.
Step 7: The length of the side CT is equal to the distance from the center of the circle to the point P, which is h plus the radius of the circle (h + r).
Step 8: Expressing the Pythagorean theorem in terms of PT, we have (h + r)^2 = d^2 + r^2.
Step 9: Expand and simplify the equation: h^2 + 2hr + r^2 = d^2 + r^2.
Step 10: Subtract r^2 from both sides of the equation: h^2 + 2hr = d^2.
Step 11: Finally, take the square root of both sides of the equation to solve for d: d = √(h^2 + 2hr).
Therefore, the distance from the point P to the point of tangency T, denoted by d, is equal to the square root of (h^2 + 2hr).
To derive a function for d in terms of h, we can use the properties of right triangles and circles. Let's denote the center of the circle as C.
Since PT is perpendicular to CT, we have a right triangle PTC. The radius r of the circle is the hypotenuse of this triangle.
We can apply the Pythagorean theorem to relate the sides of the right triangle:
PT^2 + CT^2 = PC^2.
Now, let's focus on the lengths involved:
- PT: This is the distance we're trying to express as a function of h.
- CT: This is the radius of the circle, which is r.
- PC: This is the distance from P to C. To find this, we can use the fact that CT is perpendicular to PT. The triangle PTC is similar to the right triangle PHC, where PH is the height of the triangle. Since the radius is perpendicular to the tangent line at the point of tangency, PH is the perpendicular distance from P to the line through C. So, we can find PC using the Pythagorean theorem in triangle PHC:
PH^2 + HC^2 = PC^2.
Consider right triangle PHC:
- HC: This is equal to r since it is the radius of the circle.
- PH: This is equal to h since it is the perpendicular distance from P to the tangent line at T.
Substituting these values, we have:
h^2 + r^2 = PC^2.
Now, let's substitute PC^2 into our previous equation for PT^2 + CT^2:
PT^2 + CT^2 = h^2 + r^2.
Since PT = d and CT = r, we can rewrite this as:
d^2 + r^2 = h^2 + r^2.
Notice that the r^2 terms on both sides cancel out:
d^2 = h^2.
Taking the square root of both sides, we have:
d = √(h^2).
Simplifying:
d = |h|.
Therefore, the distance d from the point P to the point of tangency T can be expressed as a function of h as:
d(h) = |h|.