14. The radius of circle 0 is 3 and MN = 5. Part A: MP is tangent to circle 0 at point P. How are MP and OP related? How do you know? Part B: Describe how you could find MP. Part C: Find MP. Round to the nearest tenth if necessary. M
14. The radius of circle 0 is 3 and MN = 5. Part A: MP is tangent to circle 0 at point P. How are MP and OP related? How do you know? Part B: Describe how you could find MP. Part C: Find MP. Round to the nearest tenth if necessary. M
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![**Problem 14: Geometry Question**
**Given Information:**
- The radius of circle \(O\) is 3.
- \(MN = 5\).
**Part A:**
\(MP\) is tangent to circle \(O\) at point \(P\). How are \(\overline{MP}\) and \(\overline{OP}\) related? How do you know?
**Part B:**
Describe how you could find \(MP\).
**Part C:**
Find \(MP\). Round to the nearest tenth if necessary.
**Diagram Explanation:**
- The diagram depicts a circle with center \(O\) and radius \(3\).
- Point \(P\) lies on the circle.
- \(MN\) is a line segment with \(N\) on the circle and \(M\) outside the circle.
- Line segment \(MP\) is a tangent to the circle at point \(P\).
- Line segment \(\overline{OP}\) is the radius of the circle from the center \(O\) to the tangent point \(P\).
**Part A: Detailed Explanation**
In geometry, a tangent to a circle is perpendicular to the radius at the point of tangency. Therefore:
- \(\overline{MP}\) (the tangent) is perpendicular to \(\overline{OP}\) (the radius).
- We know this because, by definition, a tangent to a circle always forms a 90-degree angle with the radius at the point of tangency.
**Part B: Detailed Explanation**
To find \(MP\), we can use the Pythagorean Theorem in the right triangle \(MOP\), where:
- \(\overline{OP}\) is the radius = 3 units,
- \(MN\) is given as 5 units (which is the hypotenuse of \(\triangle MON\)),
- \(\overline{ON}\) (being part of the radius) also equals 3 units.
We recognize \(\overline{MO} = \overline{MP}\) because \(M, P, O\) form a right triangle with \(\overline{OP}\) as one side.
Using the Pythagorean Theorem in \(\triangle MOP\):
\[
MO^2 = MP^2 + OP^2
\]
Since \(MO\) equals \(MN - ON\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c8a549a-ce13-4c73-8e46-ff62e8d935c2%2Ffa4662a1-5ed6-4a77-a06c-c59202a0752a%2Fpb69l2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 14: Geometry Question**
**Given Information:**
- The radius of circle \(O\) is 3.
- \(MN = 5\).
**Part A:**
\(MP\) is tangent to circle \(O\) at point \(P\). How are \(\overline{MP}\) and \(\overline{OP}\) related? How do you know?
**Part B:**
Describe how you could find \(MP\).
**Part C:**
Find \(MP\). Round to the nearest tenth if necessary.
**Diagram Explanation:**
- The diagram depicts a circle with center \(O\) and radius \(3\).
- Point \(P\) lies on the circle.
- \(MN\) is a line segment with \(N\) on the circle and \(M\) outside the circle.
- Line segment \(MP\) is a tangent to the circle at point \(P\).
- Line segment \(\overline{OP}\) is the radius of the circle from the center \(O\) to the tangent point \(P\).
**Part A: Detailed Explanation**
In geometry, a tangent to a circle is perpendicular to the radius at the point of tangency. Therefore:
- \(\overline{MP}\) (the tangent) is perpendicular to \(\overline{OP}\) (the radius).
- We know this because, by definition, a tangent to a circle always forms a 90-degree angle with the radius at the point of tangency.
**Part B: Detailed Explanation**
To find \(MP\), we can use the Pythagorean Theorem in the right triangle \(MOP\), where:
- \(\overline{OP}\) is the radius = 3 units,
- \(MN\) is given as 5 units (which is the hypotenuse of \(\triangle MON\)),
- \(\overline{ON}\) (being part of the radius) also equals 3 units.
We recognize \(\overline{MO} = \overline{MP}\) because \(M, P, O\) form a right triangle with \(\overline{OP}\) as one side.
Using the Pythagorean Theorem in \(\triangle MOP\):
\[
MO^2 = MP^2 + OP^2
\]
Since \(MO\) equals \(MN - ON\
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