Let C 1 and C 2 be two circles which do not externally touch and intersect each other and O 1 , and O 2 be the

Question

Let C 1 and C 2 be two circles which do not externally touch and intersect each other and O 1 , and O 2 be the centers of the circles, respectively. Let AB be the common transverse tangent to the circles such that P, Q are the points of tangency respectively to C 1 , C 2 . Let R be the point of intersection of O 1 O 2 and AB. If ∠ PO 1 R = 60°, find ∠ QO 2 R and ∠ QRO 2 , respectively.

Examoogle · Accepted Answer

The correct answer is Option D: $60^\circ$ and $30^\circ$. In the case of common transverse tangents, the triangles formed by centers, tangency points, and the intersection point on the line of centers are similar, thus corresponding angles are equal.

Step-by-Step Solution & Concepts:
Shortcut Trick: In the case of common transverse tangents, the triangles formed by centers, tangency points, and the intersection point on the line of centers are similar, thus corresponding angles are equal.

Formula Used: $$\Delta O_1PR \sim \Delta O_2QR$$

Given: Two circles $C_1, C_2$ with centers $O_1, O_2$ and common transverse tangent $AB$ touching at $P$ and $Q$. Transverse tangent $AB$ meets $O_1O_2$ at $R$. Given $\angle PO_1R = 60^\circ$.

Steps: Analyze Tangency Properties: Since $AB$ is a tangent to $C_1$ at $P$, $O_1P \perp AB$. Similarly, $O_2Q \perp AB$. Thus, $\angle O_1PR = \angle O_2QR = 90^\circ$.. Establish Similarity: Triangles $\Delta O_1PR$ and $\Delta O_2QR$ share vertically opposite angles at $R$ ($\angle O_1RP = \angle O_2RQ$). Thus, $\Delta O_1PR \sim \Delta O_2QR$ by AA similarity.. Calculate Angles: Given $\angle PO_1R = 60^\circ$, in right $\Delta O_1PR$, $\angle PRO_1 = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. Since $\Delta O_1PR \sim \Delta O_2QR$, $\angle QO_2R = \angle PO_1R = 60^\circ$ and $\angle QRO_2 = \angle PRO_1 = 30^\circ$.

Answer: D is correct because the similarity of the two right-angled triangles ensures that corresponding angles match the given $60^\circ$ and the derived $30^\circ$.

Exam Tip: This property of transverse tangents forming similar triangles is analogous to properties of common direct tangents and helps in solving complex geometry coordinate problems.

Examoogle · Answer

Option A: $45^\circ$ and $45^\circ$

Examoogle · Answer

Option B: $20^\circ$ and $70^\circ$

Examoogle · Answer

Option C: $40^\circ$ and $50^\circ$

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Discussion (0)

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)