Question:
If PQ and PR are tangents to the circle with centre O and radius 4 cm such that \(\angle QPR = 90^\circ\), then the length OP is
If PQ and PR are tangents to the circle with centre O and radius 4 cm such that \(\angle QPR = 90^\circ\), then the length OP is
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Whenever the angle between two tangents drawn from an external point to a circle of radius \(r\) is \(90^\circ\), the quadrilateral formed by the radii and the tangents is always a square.
The distance from the center to the external point is simply the diagonal of this square, which is calculated as \(r\sqrt{2}\).
Recognizing this immediately allows you to bypass the calculation!
The distance from the center to the external point is simply the diagonal of this square, which is calculated as \(r\sqrt{2}\).
Recognizing this immediately allows you to bypass the calculation!
Updated On: Jul 9, 2026
- 4 cm
- \(4\sqrt{2}\) cm
- 8 cm
- \(2\sqrt{2}\) cm
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The topic of this question is Circles and Tangents.
A primary theorem of circle geometry states that the radius of a circle is perpendicular to the tangent at the point of contact.
We are given a circle with center \(O\) and radius \(OQ = OR = 4\) cm.
Two tangents \(PQ\) and \(PR\) are drawn from an external point \(P\).
The angle between the two tangents is \(\angle QPR = 90^\circ\).
We need to calculate the length of the line segment \(OP\) joining the center of the circle to the external point \(P\).
Step 2: Key Formula or Approach:
- Since \(PQ\) and \(PR\) are tangents, we know that \(\angle OQP = 90^\circ\) and \(\angle ORP = 90^\circ\).
- We can analyze the angles of the quadrilateral \(OQPR\) to determine its geometric properties.
- We can apply the Pythagoras theorem in the right-angled triangle \(\Delta OQP\) to find the hypotenuse \(OP\):
\[ OP^2 = OQ^2 + PQ^2 \]
Step 3: Detailed Explanation:
• Consider the quadrilateral \(OQPR\) formed by the radii and the tangents:
The sum of the interior angles of any quadrilateral is \(360^\circ\).
We are given:
\[ \angle QPR = 90^\circ \] From the properties of tangents and radii, we have:
\[ \angle OQP = 90^\circ \quad \text{and} \quad \angle ORP = 90^\circ \]
• Find the remaining angle \(\angle QOR\):
\[ \angle QOR = 360^\circ - (\angle QPR + \angle OQP + \angle ORP) \] \[ \angle QOR = 360^\circ - (90^\circ + 90^\circ + 90^\circ) = 90^\circ \]
• Analyze the shape of the quadrilateral:
Since all four interior angles are exactly \(90^\circ\), \(OQPR\) is a rectangle.
Furthermore, the adjacent sides are equal because they are both radii of the same circle:
\[ OQ = OR = 4 \text{ cm} \] A rectangle with equal adjacent sides is a square.
Therefore, the quadrilateral \(OQPR\) is a square, which implies:
\[ PQ = PR = 4 \text{ cm} \]
• Apply the Pythagoras theorem to the right-angled triangle \(\Delta OQP\):
\[ OP^2 = OQ^2 + PQ^2 \] \[ OP^2 = 4^2 + 4^2 \] \[ OP^2 = 16 + 16 = 32 \] \[ OP = \sqrt{32} = 4\sqrt{2} \text{ cm} \]
Step 4: Final Answer:
The length of \(OP\) is \(4\sqrt{2}\) cm.
Therefore, the correct option is (B).
The topic of this question is Circles and Tangents.
A primary theorem of circle geometry states that the radius of a circle is perpendicular to the tangent at the point of contact.
We are given a circle with center \(O\) and radius \(OQ = OR = 4\) cm.
Two tangents \(PQ\) and \(PR\) are drawn from an external point \(P\).
The angle between the two tangents is \(\angle QPR = 90^\circ\).
We need to calculate the length of the line segment \(OP\) joining the center of the circle to the external point \(P\).
Step 2: Key Formula or Approach:
- Since \(PQ\) and \(PR\) are tangents, we know that \(\angle OQP = 90^\circ\) and \(\angle ORP = 90^\circ\).
- We can analyze the angles of the quadrilateral \(OQPR\) to determine its geometric properties.
- We can apply the Pythagoras theorem in the right-angled triangle \(\Delta OQP\) to find the hypotenuse \(OP\):
\[ OP^2 = OQ^2 + PQ^2 \]
Step 3: Detailed Explanation:
• Consider the quadrilateral \(OQPR\) formed by the radii and the tangents:
The sum of the interior angles of any quadrilateral is \(360^\circ\).
We are given:
\[ \angle QPR = 90^\circ \] From the properties of tangents and radii, we have:
\[ \angle OQP = 90^\circ \quad \text{and} \quad \angle ORP = 90^\circ \]
• Find the remaining angle \(\angle QOR\):
\[ \angle QOR = 360^\circ - (\angle QPR + \angle OQP + \angle ORP) \] \[ \angle QOR = 360^\circ - (90^\circ + 90^\circ + 90^\circ) = 90^\circ \]
• Analyze the shape of the quadrilateral:
Since all four interior angles are exactly \(90^\circ\), \(OQPR\) is a rectangle.
Furthermore, the adjacent sides are equal because they are both radii of the same circle:
\[ OQ = OR = 4 \text{ cm} \] A rectangle with equal adjacent sides is a square.
Therefore, the quadrilateral \(OQPR\) is a square, which implies:
\[ PQ = PR = 4 \text{ cm} \]
• Apply the Pythagoras theorem to the right-angled triangle \(\Delta OQP\):
\[ OP^2 = OQ^2 + PQ^2 \] \[ OP^2 = 4^2 + 4^2 \] \[ OP^2 = 16 + 16 = 32 \] \[ OP = \sqrt{32} = 4\sqrt{2} \text{ cm} \]
Step 4: Final Answer:
The length of \(OP\) is \(4\sqrt{2}\) cm.
Therefore, the correct option is (B).
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