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Two circles touch each other externally at P. QR is a common tangent to the circles touching them at Q and R. The measure of ∠QPR is
- a)45°
- b)60°
- c)90°
- d)75°
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two circles touch each other externally at P. QR is a common tangent t...
In triangle OPQ,
OP = OQ (radius of circle)
∴ ∠OQP = ∠OPQ
Let, ∠OQP = ∠OPQ = x
In triangle PCR,
CP = CR (radius of circle)
∴ ∠CPR = ∠CRP
Let, ∠CPR = ∠CRP = y
In given figure,
∠OPQ + ∠QPR + ∠RPC = 180°
⇒ x + ∠QPR + y = 180°
⇒ ∠QPR = 180° - x – y ……….eq (1)
As, we know that, radius and tangent to the circle are perpendicular to each other,
Therefore,
∠OQR and ∠CRQ are right angles
∴ ∠PQO + ∠PQR = 90°
⇒ ∠PQR = 90° - x° ………….eq (2)
∴ ∠ CRP + ∠ PRQ = 90°
⇒ ∠PRQ = 90° - y° ………..eq(3)
In triangle QPR
∠QPR + ∠PQR + ∠PRQ = 180°
⇒ 180° - x – y + 90° - x° +90° - y° = 180°
⇒ 360° - 2(x + y) = 180°
⇒ 2(x + y) = 180°
⇒ (x + y) = 90°
∠QPR = 180° - x – y
= 180° - 90°
= 90°
Most Upvoted Answer
Two circles touch each other externally at P. QR is a common tangent t...
Angle PQR is 90 degrees.
To prove this, let's draw a diagram:
First, draw the two circles with centers O1 and O2. Let the radius of the first circle be r1 and the radius of the second circle be r2. Let P be the point where the two circles touch externally.
Next, draw the tangent line QR that touches both circles at Q and R. Let Q be on the first circle and R be on the second circle.
Since QR is tangent to both circles, we know that the line segment PQ is perpendicular to QR at point Q, and the line segment PR is perpendicular to QR at point R. This is because the tangent line is always perpendicular to the radius of the circle at the point of tangency.
Now, let's look at the triangle PQR. We know that PQ is perpendicular to QR, so angle PQR is a right angle. Therefore, the measure of angle PQR is 90 degrees.
To prove this, let's draw a diagram:
First, draw the two circles with centers O1 and O2. Let the radius of the first circle be r1 and the radius of the second circle be r2. Let P be the point where the two circles touch externally.
Next, draw the tangent line QR that touches both circles at Q and R. Let Q be on the first circle and R be on the second circle.
Since QR is tangent to both circles, we know that the line segment PQ is perpendicular to QR at point Q, and the line segment PR is perpendicular to QR at point R. This is because the tangent line is always perpendicular to the radius of the circle at the point of tangency.
Now, let's look at the triangle PQR. We know that PQ is perpendicular to QR, so angle PQR is a right angle. Therefore, the measure of angle PQR is 90 degrees.
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Two circles touch each other externally at P. QR is a common tangent to the circles touching them at Q and R. The measure of ∠QPR isa)45°b)60°c)90°d)75°Correct answer is option 'C'. Can you explain this answer?
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