Conic Sections JEE 2026: Hard Circle PYQ Solved Stepwise

conic sections jee 2026: Hard Circle PYQ

The conic sections jee 2026 hard circle PYQ has the answer 80. Solve it as a four-link chain: triangle area gives the radius, the radius fixes the centre parameters, the quadrant condition selects the valid centre, and the centre-to-line distance gives the chord length. Confusing the last two lengths produces the listed wrong value 64.

This is a hard, difficulty-tag-4 Mathematics question from JEE Main 2026, 6 April morning slot. Based on its difficulty tag and four required calculation steps, its expected solving time is 120 seconds.

The circle is

x^{2} + y^{2} + 2 g x + 2 f y + 25 = 0

Its centre lies in the first quadrant and on $2 x - y = 4$

An equilateral triangle inscribed in the circle has area $27 \sqrt{3}$

You must find the square of the length of the chord cut by the vertical line $x = 1$

The supplied answer values are:

Circle of radius 6 centred at C(5, 6), with vertical chord PQ on x = 1 and perpendicular distance CM = 4

How do you find the radius in Worked Solution Part 1?

Start with the equilateral triangle because its area directly gives the circle’s radius through the official circumradius relation. First express the centre and apply the given line condition. This order gives one equation between the centre parameters before the radius supplies the second equation.

For the circle

x^{2} + y^{2} + 2 g x + 2 f y + 25 = 0

the centre is $(- g, - f)$

Substitute the centre in $2 x - y = 4$ to get $2 (- g) - (- f) = 4$

Therefore, $- 2 g + f = 4$ $f = 4 + 2 g$

For an equilateral triangle inscribed in a circle, use the official relation

Area = \frac{3 \sqrt{3}}{4} R^{2}

Substitute the given area:

\frac{3 \sqrt{3}}{4} R^{2} = 27 \sqrt{3}

Cancel the common factor:

\frac{3}{4} R^{2} = 27

Hence, $R^{2} = 36$

Since a radius is positive, $R = 6$

Converting the area to side length first adds steps. The direct area-radius relation is the faster route.

How do you locate the centre and calculate the chord in Worked Solution Part 2?

Use the radius formula with the line condition to find both possible centres. The first-quadrant condition then rejects one of them. For the valid centre, calculate its perpendicular distance from the chord’s line and use the right triangle formed by the radius, distance and half-chord.

For a circle in the form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

the radius satisfies $R^{2} = g^{2} + f^{2} - c$

Here, $c = 25$

Using the radius found above, $g^{2} + f^{2} - 25 = 36$

Therefore, $g^{2} + f^{2} = 61$

Insert $f = 4 + 2 g$ to obtain $g^{2} + (4 + 2 g)^{2} = 61$

Expand and simplify: $g^{2} + 16 + 16 g + 4 g^{2} = 61$ $5 g^{2} + 16 g - 45 = 0$

The discriminant is $D = 16^{2} - 4 \times 5 \times (- 45)$ $D = 256 + 900 = 1156 = 34^{2}$

Thus,

g = \frac{- 16 \pm 34}{10}

The two roots are $g = \frac{9}{5}$ and $g = - 5$

For the first root, $g = \frac{9}{5}$ and

f = 4 + 2 (\frac{9}{5}) = \frac{38}{5}

The corresponding centre is

(- \frac{9}{5}, - \frac{38}{5})

This centre is not in the first quadrant, so reject this root.

For the second root, $g = - 5$ and $f = 4 + 2 (- 5) = - 6$

Therefore, the valid centre is $(- g, - f) = (5, 6)$

The perpendicular distance from the valid centre to the vertical line is $d = | 5 - 1 | = 4$

The half-chord length is $\sqrt{R^{2} - d^{2}}$ $= \sqrt{36 - 16}$ $= \sqrt{20}$

Therefore, the full chord length is $2 \sqrt{20} = 4 \sqrt{5}$

Its square is $(4 \sqrt{5})^{2} = 80$

Final answer: 80

Why does the wrong answer 64 appear?

The value 64 comes from treating the perpendicular distance from the centre to the chord as the half-chord. Here the distance is 4, but it is one perpendicular leg of a right triangle. The half-chord is the other leg, while the radius is the hypotenuse.

The incorrect step is $Chord = 2 d = 2 (4) = 8$ which gives ${Chord}^{2} = 8^{2} = 64$

The centre-to-line distance is perpendicular to the chord. It is not the half-chord.

The correct Pythagorean relation is

(half-chord)^{2} + d^{2} = R^{2}

Substitute the radius and distance: $(half-chord)^{2} + 16 = 36$

(half-chord)^{2} = 36 - 16 = 20

Since the full chord is twice the half-chord, its square is ${Chord}^{2} = 4 \times 20 = 80$

For conic sections jee 2026 chord questions, write the Pythagorean relation before inserting numbers. It separates the centre-to-line distance from the half-chord.

Which related circle questions should you practise next?

Practise these three short follow-up drills from Circle, Centre, Radius and Chords. The first tests conversion from a standard circle equation. The second uses the equilateral-triangle area relation. The third gives the general result for any vertical chord, without adding unrelated conic-section theory.

How do you find the chord cut by the vertical line in the first drill?

The square of the chord is 128. Read the centre and radius from the circle equation, calculate the horizontal distance from the centre to the given vertical line, and apply the chord-square formula. This drill uses the same centre-radius-distance chain without a quadrant condition.

For

x^{2} + y^{2} - 6 x + 8 y - 11 = 0

find the square of the chord cut by $x = 1$

Answer check: $Centre = (3, - 4)$

R^{2} = 3^{2} + (- 4)^{2} - (- 11) = 36

$d = | 3 - 1 | = 2$ Therefore,

L^{2} = 4 (R^{2} - d^{2}) = 4 (36 - 4) = 128

How do you solve the equilateral-triangle variation?

The square of the chord is 48. Use the triangle’s area to find the radius squared directly. Then subtract the square of the given centre-to-chord distance and multiply the result by four, because the full chord is twice the half-chord.

An equilateral triangle of area $12 \sqrt{3}$ is inscribed in a circle. Find the square of a chord whose distance from the centre is 2.

Using the area relation,

\frac{3 \sqrt{3}}{4} R^{2} = 12 \sqrt{3}

$R^{2} = 16$ Therefore, $L^{2} = 4 (16 - 4) = 48$

What is the general formula for a chord cut by a vertical line?

The chord-square formula depends only on the radius and the horizontal distance between the centre and the vertical line. For a circle with centre coordinates and radius given below, first calculate that distance. A real chord exists only when the distance does not exceed the radius.

For a circle with centre $(h, k)$ radius $R$ and vertical line $x = a$ the perpendicular distance is $d = | h - a |$

Hence, $L^{2} = 4 (R^{2} - d^{2})$ or

L^{2} = 4 [R^{2} - (h - a)^{2}]

This chord exists provided $| h - a | \leq R$

Use the free past-paper archive to search Circle questions by chapter and solve more centre-radius-chord problems with worked solutions. For each one, write the chord-square relation before checking the answer values: $L^{2} = 4 (R^{2} - d^{2})$

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the answer to the Conic Sections JEE 2026 hard circle PYQ?

The square of the chord length is 80. The circle has radius 6 and valid centre (5, 6), whose distance from the line x = 1 is 4.

How do I find the circle radius from an inscribed equilateral triangle?

Use the relation Area = (3√3/4)R². Substituting 27√3 gives R² = 36, so the radius is 6.

What is the chord formula for a vertical line?

For a circle with centre (h, k), radius R and vertical line x = a, the perpendicular distance is |h − a|. The chord-length square is L² = 4[R² − (h − a)²].

Why is 64 the wrong answer in this circle question?

The value 64 comes from incorrectly treating the centre-to-line distance 4 as the half-chord. The correct half-chord square is 6² − 4² = 20, so the full chord square is 4 × 20 = 80.

Conic Sections JEE 2026: Hard Circle PYQ Solved