JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Area of a Triangle Formed by Two Tangents to a Circle

Q: Why can we treat PQ as the base of the triangle?

Both P and Q have the exact same x-coordinate, 1+√(5). This means the segment PQ is parallel to the y-axis. Its length is simply the difference between the y-coordinates. The perpendicular height from O is just the x-coordinate.

If the tangents drawn at the point O(0,0) and P(1+√(5), 2) on the circle x^2+y^2-2 x-4 y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

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What feels right?

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A(3+√(5))/2B(4+2 √(5))/2C(5+3 √(5))/2D(7+3 √(5))/2

Step-by-step solutionView

Correct answer

The tangents intersect at $Q (\frac{1 + \sqrt{5}}{2}, - \frac{1 + \sqrt{5}}{4})$ , giving the area of triangle $O P Q$ as $\frac{5 + 3 \sqrt{5}}{2}$ .

Option analysis

Why each option works or fails

A · (3+√(5))/2

Incorrectly omitting the cross-product terms when computing the determinant area $\frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣$ . Ensure you correctly substitute both coordinates into $\frac{1}{2} ∣ x_{1} y_{2} - x_{2} y_{1} ∣$ without dropping intermediate terms involving $\sqrt{5}$ .

B · (4+2 √(5))/2

Making an algebraic slip when expanding $(1 + \sqrt{5})^{2}$ , writing it as $1 + 5 = 6$ or $4 + 2 \sqrt{5}$ without simplifying the whole coordinate expression. Expand $(1 + \sqrt{5})^{2}$ carefully as $1 + 2 \sqrt{5} + 5 = 6 + 2 \sqrt{5}$ .

C · (5+3 √(5))/2

This is the correct option. Evaluating the intersection $Q$ of the tangent lines $x + 2 y = 0$ and $\sqrt{5} x = 1 + \sqrt{5}$ directly yields $Area (△ O P Q) = \frac{5 + 3 \sqrt{5}}{2}$ .

D · (7+3 √(5))/2

Adding rather than subtracting or making an arithmetic sign slip in computing the cross-product determinants. Check the signs carefully when taking the determinant $\frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣$ to ensure the area evaluates to $\frac{5 + 3 \sqrt{5}}{2}$ .

Reviewed route

Solution

StepWorking

01Given

Circle: $x^{2} + y^{2} - 2 x - 4 y = 0$ . Points on the circle: $O (0, 0)$ and $P (1 + \sqrt{5}, 2)$ . Tangents at $O$ and $P$ meet at $Q$ .

02Approach

First, find the equations of the tangents at $O (0, 0)$ and $P (1 + \sqrt{5}, 2)$ using $T = 0$ . Solve them simultaneously to find the intersection point $Q$ . Then, find the tangent length $L = O Q$ and circle radius $R$ , and use the area of the triangle formed by two tangents and the chord of contact: $Area (△ O P Q) = \frac{R L^{3}}{R^{2} + L^{2}}$ , or find coordinates of $Q$ and calculate $\frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣$ .

03Execute

Tangent at $O (0, 0)$ : Using $T = 0$ , $x (0) + y (0) - (x + 0) - 2 (y + 0) = 0 ⟹ x + 2 y = 0$ , so $x = - 2 y$ . Tangent at $P (1 + \sqrt{5}, 2)$ : $x (1 + \sqrt{5}) + 2 y - (x + 1 + \sqrt{5}) - 2 (y + 2) = 0 ⟹ \sqrt{5} x - (5 + \sqrt{5}) = 0$ , which gives $x = \frac{5 + \sqrt{5}}{\sqrt{5}} = 1 + \sqrt{5}$ . Since $x = - 2 y$ , we have $y = - \frac{1 + \sqrt{5}}{2}$ . Thus $Q = (1 + \sqrt{5}, - \frac{1 + \sqrt{5}}{2})$ .

04Execute

Since $O$ is $(0, 0)$ , $P$ is $(1 + \sqrt{5}, 2)$ , and $Q$ is $(1 + \sqrt{5}, - \frac{1 + \sqrt{5}}{2})$ , points $P$ and $Q$ share the same x-coordinate $x = 1 + \sqrt{5}$ . Thus, segment $P Q$ is vertical with length: $P Q = y_{P} - y_{Q} = 2 - (- \frac{1 + \sqrt{5}}{2}) = \frac{5 + \sqrt{5}}{2}$ . The horizontal distance from $O (0, 0)$ to the vertical line $x = 1 + \sqrt{5}$ is $h = 1 + \sqrt{5}$ . Therefore, the area of $△ O P Q = \frac{1}{2} \times P Q \times h = \frac{1}{2} (\frac{5 + \sqrt{5}}{2}) (1 + \sqrt{5}) = \frac{5 + 5 \sqrt{5} + \sqrt{5} + 5}{4} = \frac{10 + 6 \sqrt{5}}{4} = \frac{5 + 3 \sqrt{5}}{2}$ .

✓Verify

Alternatively, evaluate using $\frac{R L^{3}}{R^{2} + L^{2}}$ : Center $C = (1, 2)$ , $R = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$ . Length $L = O Q = \sqrt{(1 + \sqrt{5})^{2} + {(\frac{1 + \sqrt{5}}{2})}^{2}} = (1 + \sqrt{5}) \frac{\sqrt{5}}{2} = \frac{5 + \sqrt{5}}{2}$ . Then $L^{2} = \frac{30 + 10 \sqrt{5}}{4} = \frac{15 + 5 \sqrt{5}}{2}$ . $R^{2} + L^{2} = 5 + \frac{15 + 5 \sqrt{5}}{2} = \frac{25 + 5 \sqrt{5}}{2} = \sqrt{5} L$ . Hence $\frac{R L^{3}}{R^{2} + L^{2}} = \frac{\sqrt{5} L^{3}}{\sqrt{5} L} = L^{2} = \frac{15 + 5 \sqrt{5}}{2} \approx 13.09$ ? Wait: $R^{2} + L^{2} = \frac{25 + 5 \sqrt{5}}{2}$ , while $\sqrt{5} L = \frac{5 \sqrt{5} + 5}{2} \neq R^{2} + L^{2}$ . Direct calculation yields: $\frac{\sqrt{5} L^{3}}{R^{2} + L^{2}} = \frac{5 + 3 \sqrt{5}}{2}$ , exactly matching the vertical-base result.

Hints that build this answer step by step

What are the equations of the tangents to the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ at $O (0, 0)$ and $P (1 + \sqrt{5}, 2)$ ?
At $O$ : $x + 2 y = 0$ ; at $P$ : $\sqrt{5} x = 1 + \sqrt{5}$
What are the coordinates of the intersection point $Q$ of the two tangents $x + 2 y = 0$ and $\sqrt{5} x = 5 + \sqrt{5}$ ?
$Q (1 + \sqrt{5}, - \frac{1 + \sqrt{5}}{2})$
Given vertices $O (0, 0)$ , $P (1 + \sqrt{5}, 2)$ , and $Q (1 + \sqrt{5}, - \frac{1 + \sqrt{5}}{2})$ , what is the area of $△ O P Q$ ?
$\frac{5 + 3 \sqrt{5}}{2}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why can we treat PQ as the base of the triangle?

Both P and Q have the exact same x-coordinate, $1 + \sqrt{5}$ . This means the segment PQ is parallel to the y-axis. Its length is simply the difference between the y-coordinates. The perpendicular height from O is just the x-coordinate.

Answer

The tangents intersect at $Q (\frac{1 + \sqrt{5}}{2}, - \frac{1 + \sqrt{5}}{4})$ , giving the area of triangle $O P Q$ as $\frac{5 + 3 \sqrt{5}}{2}$ .

Why each option works or fails

A: (3+√(5))/2 - Incorrectly omitting the cross-product terms when computing the determinant area $\frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣$ . Ensure you correctly substitute both coordinates into $\frac{1}{2} ∣ x_{1} y_{2} - x_{2} y_{1} ∣$ without dropping intermediate terms involving $\sqrt{5}$ .
B: (4+2 √(5))/2 - Making an algebraic slip when expanding $(1 + \sqrt{5})^{2}$ , writing it as $1 + 5 = 6$ or $4 + 2 \sqrt{5}$ without simplifying the whole coordinate expression. Expand $(1 + \sqrt{5})^{2}$ carefully as $1 + 2 \sqrt{5} + 5 = 6 + 2 \sqrt{5}$ .
C · correct: (5+3 √(5))/2 - This is the correct option. Evaluating the intersection $Q$ of the tangent lines $x + 2 y = 0$ and $\sqrt{5} x = 1 + \sqrt{5}$ directly yields $Area (△ O P Q) = \frac{5 + 3 \sqrt{5}}{2}$ .
D: (7+3 √(5))/2 - Adding rather than subtracting or making an arithmetic sign slip in computing the cross-product determinants. Check the signs carefully when taking the determinant $\frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣$ to ensure the area evaluates to $\frac{5 + 3 \sqrt{5}}{2}$ .

Step-by-step solution

Given: Circle: $x^{2} + y^{2} - 2 x - 4 y = 0$ . Points on the circle: $O (0, 0)$ and $P (1 + \sqrt{5}, 2)$ . Tangents at $O$ and $P$ meet at $Q$ .
Approach: First, find the equations of the tangents at $O (0, 0)$ and $P (1 + \sqrt{5}, 2)$ using $\sqrt{5} x = 1 + \sqrt{5}$ . Solve them simultaneously to find the intersection point $Q$ . Then, find the tangent length $L = O Q$ and circle radius $R$ , and use the area of the triangle formed by two tangents and the chord of contact: $Area (△ O P Q) = \frac{R L^{3}}{R^{2} + L^{2}}$ , or find coordinates of $Q$ and calculate $\frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣$ .
Execute: Tangent at $O (0, 0)$ : Using $T = 0$ , $x (0) + y (0) - (x + 0) - 2 (y + 0) = 0 ⟹ x + 2 y = 0$ , so $x = - 2 y$ . Tangent at $P (1 + \sqrt{5}, 2)$ : $x (1 + \sqrt{5}) + 2 y - (x + 1 + \sqrt{5}) - 2 (y + 2) = 0 ⟹ \sqrt{5} x - (5 + \sqrt{5}) = 0$ , which gives $x = \frac{5 + \sqrt{5}}{\sqrt{5}} = 1 + \sqrt{5}$ . Since $x = - 2 y$ , we have $y = - \frac{1 + \sqrt{5}}{2}$ . Thus $Q = (1 + \sqrt{5}, - \frac{1 + \sqrt{5}}{2})$ .
Execute: Since $O$ is $(0, 0)$ , $P$ is $(1 + \sqrt{5}, 2)$ , and $Q$ is $(1 + \sqrt{5}, - \frac{1 + \sqrt{5}}{2})$ , points $P$ and $Q$ share the same x-coordinate $x = 1 + \sqrt{5}$ . Thus, segment $P Q$ is vertical with length: $P Q = y_{P} - y_{Q} = 2 - (- \frac{1 + \sqrt{5}}{2}) = \frac{5 + \sqrt{5}}{2}$ . The horizontal distance from $O (0, 0)$ to the vertical line $x = 1 + \sqrt{5}$ is $h = 1 + \sqrt{5}$ . Therefore, the area of $△ O P Q = \frac{1}{2} \times P Q \times h = \frac{1}{2} (\frac{5 + \sqrt{5}}{2}) (1 + \sqrt{5}) = \frac{5 + 5 \sqrt{5} + \sqrt{5} + 5}{4} = \frac{10 + 6 \sqrt{5}}{4} = \frac{5 + 3 \sqrt{5}}{2}$ .
Verify: Alternatively, evaluate using $\frac{R L^{3}}{R^{2} + L^{2}}$ : Center $C = (1, 2)$ , $O (0, 0)$ . Length $L = O Q = \sqrt{(1 + \sqrt{5})^{2} + {(\frac{1 + \sqrt{5}}{2})}^{2}} = (1 + \sqrt{5}) \frac{\sqrt{5}}{2} = \frac{5 + \sqrt{5}}{2}$ . Then $L^{2} = \frac{30 + 10 \sqrt{5}}{4} = \frac{15 + 5 \sqrt{5}}{2}$ . $R^{2} + L^{2} = 5 + \frac{15 + 5 \sqrt{5}}{2} = \frac{25 + 5 \sqrt{5}}{2} = \sqrt{5} L$ . Hence $\frac{R L^{3}}{R^{2} + L^{2}} = \frac{\sqrt{5} L^{3}}{\sqrt{5} L} = L^{2} = \frac{15 + 5 \sqrt{5}}{2} \approx 13.09$ ? Wait: $R^{2} + L^{2} = \frac{25 + 5 \sqrt{5}}{2}$ , while $\sqrt{5} L = \frac{5 \sqrt{5} + 5}{2} \neq R^{2} + L^{2}$ . Direct calculation yields: $\frac{\sqrt{5} L^{3}}{R^{2} + L^{2}} = \frac{5 + 3 \sqrt{5}}{2}$ , exactly matching the vertical-base result.

Area of a Triangle Formed by Two Tangents to a Circle

What feels right?

The tangents intersect at $Q (\frac{1 + \sqrt{5}}{2}, - \frac{1 + \sqrt{5}}{4})$ , giving the area of triangle $O P Q$ as $\frac{5 + 3 \sqrt{5}}{2}$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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