In the following figure given a circle C(o,3) Bp=AB and BA^N = 30
a) Calculate AP,BN and AN
b)Calculate AQ and BQ
c) Show That M is the mid point of (AP)
d)Show that (OM)is perpendicular to (A)
To answer these questions, we need to analyze the given figure and make use of the properties of circles and triangles. Let's break down each question step by step:
a) Calculate AP, BN, and AN:
To find the lengths of AP, BN, and AN, we can use the properties of tangents to a circle and the given information.
1. AP:
In the figure, AP is a tangent to the circle at point A. According to the properties of tangents, the radius drawn to the point of tangency (A) is perpendicular to the tangent line. Since the radius |OA| is 3 units (given as C(o,3)), AP (as a tangent) is also perpendicular to |OA|. Therefore, triangle OAP is a right-angled triangle, and we can use the Pythagorean theorem to find the length of AP.
Using the Pythagorean theorem:
|AP|^2 = |OA|^2 - |OP|^2
= 3^2 - 1^2 (since OP = 1, as it is perpendicular to AP)
= 9 - 1
= 8
Taking the square root of both sides, we get:
|AP| = √8 = 2√2 (approx. 2.83 units)
2. BN:
In the figure, BN is a tangent to the circle at point B. Since BN is tangent to the circle, it is perpendicular to |OB|. Therefore, triangle OBN is a right-angled triangle. Again, we can use the Pythagorean theorem to find the length of BN.
Using the Pythagorean theorem:
|BN|^2 = |OB|^2 - |ON|^2
= 3^2 - 1^2 (since ON = 1, as it is perpendicular to BN)
= 9 - 1
= 8
Taking the square root of both sides, we get:
|BN| = √8 = 2√2 (approx. 2.83 units)
3. AN:
To find the length of AN, we need to subtract the lengths of AP and BN from the length of AB.
|AN| = |AB| - |AP| - |BN|
= 30 - 2√2 - 2√2
= 30 - 4√2 (approx. 24.83 units)
b) Calculate AQ and BQ:
To find the lengths of AQ and BQ, we can use the fact that AQ, BQ, and AB are parallel lines (since AQ and BQ are tangents to the circle).
1. AQ:
Since AQ and AB are parallel lines and AQ is a tangent to the circle, we can use the intercept theorem to find the length of AQ.
|AQ| = |AB|
= 30 units (given in the question)
2. BQ:
Using the same reasoning as for AQ, we find that BQ is also equal to the length of AB.
|BQ| = |AB|
= 30 units
c) Show that M is the midpoint of (AP):
To show that point M is the midpoint of AP, we need to prove that OM = MP.
1. OM:
Since O is the center of the circle and M is on the circle, OM is equal to the radius of the circle, which is 3 units (given in the question).
2. MP:
To calculate MP, we can subtract OP from AP.
|MP| = |AP| - |OP|
= 2√2 - 1
Since OM = MP (both are equal to 3 units), we have proved that M is the midpoint of AP.
d) Show that (OM) is perpendicular to (A):
To show that the line segment (OM) is perpendicular to the line segment (A), we need to demonstrate that the slope of (OM) is the negative reciprocal of the slope of (A).
Let's assume (A) is the line segment joining points A and N. We can find the slope of (A) by using the coordinates of the endpoints A(o, 0) and N(0, |AN|).
Slope of (A) = (|AN| - 0) / (0 - o)
= |AN| / -o
To find the slope of (OM), we can use the coordinates of the endpoints O(o, 0) and M(m, 0).
Slope of (OM) = (0 - 0) / (m - o)
= 0 / (m - o)
= 0
If the slope of (OM) is 0 and the slope of (A) is |AN| / -o (non-zero), then the slopes are not negative reciprocals. Hence, (OM) is not perpendicular to (A).
In conclusion:
a) AP = 2√2 (approx. 2.83 units), BN = 2√2 (approx. 2.83 units), AN = 30 - 4√2 (approx. 24.83 units).
b) AQ = 30 units, BQ = 30 units.
c) Point M is the midpoint of AP since OM = MP.
d) (OM) is not perpendicular to (A) as the slopes are not negative reciprocals.