JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding the Incenter of a Triangle Formed by Three Tangent Lines

Q: Why does h + k simplify so neatly without radicals?

Because the numerator 50 + 35√(2) is an exact multiple (5×) of the denominator 10 + 7√(2).

Let y=x+2,4 y=3 x+6 and 3 y=4 x+1 be three tangent lines to the circle (x-h)^2+(y-k)^2=r^2. Then h+k is equal to:

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A5(1+√(2))B5 √(2)C6D5

Step-by-step solutionView

Correct answer

The center of the inscribed circle is $(h, k) = (2, 3)$ , which gives $h + k = 5$ .

Option analysis

Why each option works or fails

A · 5(1+√(2))

Adding the distance to a vertex or including an extraneous $\sqrt{2}$ factor from the angle bisector equations without simplifying. Solve for the point of intersection of the interior angle bisectors directly; the coordinates evaluate to integers $(2, 3)$ .

B · 5 √(2)

Confusing the coordinates with terms involving $\sqrt{1^{2} + 1^{2}} = \sqrt{2}$ from the distance formula to line $x - y + 2 = 0$ . The distance $r$ involves $\sqrt{2}$ , but the center coordinates $(h, k)$ are the intersection of the angle bisectors, which are rational numbers.

C · 6

Making an arithmetic error when solving the system of angle bisector equations, such as taking $h = 3, k = 3$ . Substitute $(h, k)$ into the equidistant relations carefully: $∣ h - k + 2 ∣ / \sqrt{2} = ∣ 3 h - 4 k + 6 ∣ / 5 = ∣ 4 h - 3 k + 1 ∣ / 5$ to verify $h = 2, k = 3$ .

D · 5

The student correctly determines the center $(h, k) = (2, 3)$ and adds $h + k = 5$ . Keep using symmetry: the lines $3 x - 4 y + 6 = 0$ and $4 x - 3 y + 1 = 0$ intersect at $(2, 3)$ , and their internal angle bisector is along the symmetry axis.

Reviewed route

Solution

StepWorking

01Given

Three tangent lines to the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ are: $L_{1} : x - y + 2 = 0$ $L_{2} : 3 x - 4 y + 6 = 0$ $L_{3} : 4 x - 3 y + 1 = 0$

02Goal

Find the value of $h + k$ , where $(h, k)$ is the centre of the incircle of the triangle formed by $L_{1}, L_{2}, L_{3}$ .

03Approach

Find the three vertices of the triangle by pairwise intersection of the lines, calculate the lengths of the opposite sides, and apply the incentre formula: $h = \frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}$ , $k = \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}$ . Then compute $h + k$ .

04Execute

Solve pairwise intersections: 1. $L_{1} \cap L_{2}$ : $y = x + 2 ⟹ 4 (x + 2) = 3 x + 6 ⟹ x = - 2, y = 0$ . Vertex $A = (- 2, 0)$ . 2. $L_{1} \cap L_{3}$ : $y = x + 2 ⟹ 3 (x + 2) = 4 x + 1 ⟹ x = 5, y = 7$ . Vertex $B = (5, 7)$ . 3. $L_{2} \cap L_{3}$ : Multiply $L_{2}$ by $3$ and $L_{3}$ by $4$ : $9 x - 12 y + 18 = 0$ and $16 x - 12 y + 4 = 0$ . Subtracting gives $7 x - 14 = 0 ⟹ x = 2$ , then $4 y = 12 ⟹ y = 3$ . Vertex $C = (2, 3)$ .

05Execute

Calculate the side lengths opposite to vertices $A, B, C$ : $a = B C = \sqrt{(2 - 5)^{2} + (3 - 7)^{2}} = \sqrt{(- 3)^{2} + (- 4)^{2}} = 5$ $b = A C = \sqrt{(2 - (- 2))^{2} + (3 - 0)^{2}} = \sqrt{4^{2} + 3^{2}} = 5$ $c = A B = \sqrt{(5 - (- 2))^{2} + (7 - 0)^{2}} = \sqrt{7^{2} + 7^{2}} = 7 \sqrt{2}$

06Execute

Apply the incentre formula for $h$ and $k$ : Perimeter $P = a + b + c = 5 + 5 + 7 \sqrt{2} = 10 + 7 \sqrt{2}$ . $h = \frac{a x_{A} + b x_{B} + c x_{C}}{a + b + c} = \frac{5 (- 2) + 5 (5) + 7 \sqrt{2} (2)}{10 + 7 \sqrt{2}} = \frac{15 + 14 \sqrt{2}}{10 + 7 \sqrt{2}}$ $k = \frac{a y_{A} + b y_{B} + c y_{C}}{a + b + c} = \frac{5 (0) + 5 (7) + 7 \sqrt{2} (3)}{10 + 7 \sqrt{2}} = \frac{35 + 21 \sqrt{2}}{10 + 7 \sqrt{2}}$ $h + k = \frac{(15 + 14 \sqrt{2}) + (35 + 21 \sqrt{2})}{10 + 7 \sqrt{2}} = \frac{50 + 35 \sqrt{2}}{10 + 7 \sqrt{2}} = \frac{5 (10 + 7 \sqrt{2})}{10 + 7 \sqrt{2}} = 5$ .

✓Verify

Check that the incentre is equidistant from all three lines: Distance from $(h, k)$ to $x - y + 2 = 0$ : $∣ h - k + 2 ∣ / \sqrt{2}$ . Notice $△ A B C$ is isosceles with $A C = B C = 5$ . The line of symmetry is the angle bisector of $∠ C$ , passing through $C (2, 3)$ and the midpoint of $A B$ , which is $(1.5, 3.5)$ . Along this line, $x + y = 5$ , which confirms $h + k = 5$ .

✓ 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: 9 September 2026

Quick checks

Students also ask

Why does

h + k

simplify so neatly without radicals?

Because the numerator $50 + 35 \sqrt{2}$ is an exact multiple ( $5 \times$ ) of the denominator $10 + 7 \sqrt{2}$ .

Does the incentre always lie on the median?

Only in an isosceles triangle does the angle bisector of the vertex angle coincide with the median to the base.

Answer

The center of the inscribed circle is $(h, k) = (2, 3)$ , which gives $h + k = 5$ .

Why each option works or fails

A: 5(1+√(2)) - Adding the distance to a vertex or including an extraneous $\sqrt{2}$ factor from the angle bisector equations without simplifying. Solve for the point of intersection of the interior angle bisectors directly; the coordinates evaluate to integers $(2, 3)$ .
B: 5 √(2) - Confusing the coordinates with terms involving $\sqrt{1^{2} + 1^{2}} = \sqrt{2}$ from the distance formula to line $x - y + 2 = 0$ . The distance $r$ involves $\sqrt{2}$ , but the center coordinates $(h, k)$ are the intersection of the angle bisectors, which are rational numbers.
C: 6 - Making an arithmetic error when solving the system of angle bisector equations, such as taking $h = 3, k = 3$ . Substitute $(h, k)$ into the equidistant relations carefully: $∣ h - k + 2 ∣ / \sqrt{2} = ∣ 3 h - 4 k + 6 ∣ / 5 = ∣ 4 h - 3 k + 1 ∣ / 5$ to verify $h = 2, k = 3$ .
D · correct: 5 - The student correctly determines the center $(h, k) = (2, 3)$ and adds $h + k = 5$ . Keep using symmetry: the lines $3 x - 4 y + 6 = 0$ and $4 x - 3 y + 1 = 0$ intersect at $(2, 3)$ , and their internal angle bisector is along the symmetry axis.

Step-by-step solution

Given: Three tangent lines to the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ are: $L_{1} : x - y + 2 = 0$ $L_{2} : 3 x - 4 y + 6 = 0$ $L_{3} : 4 x - 3 y + 1 = 0$
Goal: Find the value of $h + k$ , where $(h, k)$ is the centre of the incircle of the triangle formed by $L_{1}, L_{2}, L_{3}$ .
Approach: Find the three vertices of the triangle by pairwise intersection of the lines, calculate the lengths of the opposite sides, and apply the incentre formula: $h = \frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}$ , $k = \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}$ . Then compute $h + k$ .
Execute: Solve pairwise intersections: 1. $L_{1} \cap L_{2}$ : $y = x + 2 ⟹ 4 (x + 2) = 3 x + 6 ⟹ x = - 2, y = 0$ . Vertex $A = (- 2, 0)$ . 2. $L_{1} \cap L_{3}$ : $y = x + 2 ⟹ 3 (x + 2) = 4 x + 1 ⟹ x = 5, y = 7$ . Vertex $B = (5, 7)$ . 3. $L_{2} \cap L_{3}$ : Multiply $L_{2}$ by $3$ and $L_{3}$ by $4$ : $9 x - 12 y + 18 = 0$ and $16 x - 12 y + 4 = 0$ . Subtracting gives $7 x - 14 = 0 ⟹ x = 2$ , then $4 y = 12 ⟹ y = 3$ . Vertex $C = (2, 3)$ .
Execute: Calculate the side lengths opposite to vertices $A, B, C$ : $a = B C = \sqrt{(2 - 5)^{2} + (3 - 7)^{2}} = \sqrt{(- 3)^{2} + (- 4)^{2}} = 5$ $b = A C = \sqrt{(2 - (- 2))^{2} + (3 - 0)^{2}} = \sqrt{4^{2} + 3^{2}} = 5$ $c = A B = \sqrt{(5 - (- 2))^{2} + (7 - 0)^{2}} = \sqrt{7^{2} + 7^{2}} = 7 \sqrt{2}$
Execute: Apply the incentre formula for $h$ and $k$ : Perimeter $P = a + b + c = 5 + 5 + 7 \sqrt{2} = 10 + 7 \sqrt{2}$ . $h = \frac{a x_{A} + b x_{B} + c x_{C}}{a + b + c} = \frac{5 (- 2) + 5 (5) + 7 \sqrt{2} (2)}{10 + 7 \sqrt{2}} = \frac{15 + 14 \sqrt{2}}{10 + 7 \sqrt{2}}$ $k = \frac{a y_{A} + b y_{B} + c y_{C}}{a + b + c} = \frac{5 (0) + 5 (7) + 7 \sqrt{2} (3)}{10 + 7 \sqrt{2}} = \frac{35 + 21 \sqrt{2}}{10 + 7 \sqrt{2}}$ $h + k = \frac{(15 + 14 \sqrt{2}) + (35 + 21 \sqrt{2})}{10 + 7 \sqrt{2}} = \frac{50 + 35 \sqrt{2}}{10 + 7 \sqrt{2}} = \frac{5 (10 + 7 \sqrt{2})}{10 + 7 \sqrt{2}} = 5$ .
Verify: Check that the incentre is equidistant from all three lines: Distance from $(h, k)$ to $x - y + 2 = 0$ : $∣ h - k + 2 ∣ / \sqrt{2}$ . Notice $△ A B C$ is isosceles with $A C = B C = 5$ . The line of symmetry is the angle bisector of $∠ C$ , passing through $C (2, 3)$ and the midpoint of $A B$ , which is $(1.5, 3.5)$ . Along this line, $x + y = 5$ , which confirms $h + k = 5$ .

Shortcut:

When to use it: When two sides of the triangle are equal, the incentre lies on the axis of symmetry.

Given: Vertices of the triangle are $A (- 2, 0)$ , $B (5, 7)$ , $C (2, 3)$ , with side lengths $B C = 5$ , $A C = 5$ , and $A B = 7 \sqrt{2}$ .

Approach: Since $A C = B C = 5$ , $△ A B C$ is isosceles with apex at $C$ . The incentre $(h, k)$ must lie on the internal bisector of $∠ C$ , which is also the median to $A B$ .

Execute: The midpoint of $A B$ is $M = (\frac{- 2 + 5}{2}, \frac{0 + 7}{2}) = (\frac{3}{2}, \frac{7}{2})$ . The equation of line $C M$ passing through $C (2, 3)$ and $M (1.5, 3.5)$ has slope $h + k$ . Equation of $C M$ : $y - 3 = - 1 (x - 2) ⟹ x + y = 5$ . Since the incentre $(h, k)$ lies on this axis of symmetry, it must satisfy $h + k = 5$ .

Verify: The point $C (2, 3)$ gives $x + y = 5$ , and $M (1.5, 3.5)$ gives $1.5 + 3.5 = 5$ . Any point on the median has $x + y = 5$ , so $h + k = 5$ immediately.

Finding the Incenter of a Triangle Formed by Three Tangent Lines

What feels right?

The center of the inscribed circle is $(h, k) = (2, 3)$ , which gives $h + k = 5$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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