Let the equation of the circle, which touches x-axis at the point (a, 0), a

Year	2026	2025	2024	Total
Questions	6	6	5	17

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Circle intercepts standard form templates:

y-intercept = 2\sqrt(f² - c)

Core Logic

Since the circle touches the x-axis at $(a,0)$ $(a,0)$ and lies entirely below it, its center is located at $(a, -p)$ $(a, -p)$ where $p$ $p$ matches its radius $r$ $r$.

By Pythagoras' theorem:

Circles Touching Axes and Intercepts diagram for Q58 - JEE Main 2025 Morning

r² = a² + (b²)/(4) = p²

Step 1: Translating to General Equation Form

The explicit standard equation is $(x-a)² + (y+p)² = r²$ $(x-a)^2 + (y+p)^2 = r^2$. Expanding it out:

x² + y² - 2ax + 2py + a² = 0

Comparing this directly to $x² + y² - α x + β y + γ = 0$ yields: $α = 2a$ $\alpha = 2a$, $β = 2p$ $\beta = 2p$, and $γ = a²$ $\gamma = a^2$.

Step 2: Evaluating the Target Mapped Ordered Pair

Isolating $b²$ $b^2$ using the parametric radius dimensions:

b² = 4p² - 4a² = (2p)² - 4(a²) = β² - 4γ

Thus, the mapped ordered pair $(2a, b²)$ $(2a, b^2)$ evaluates directly to $(α, β² - 4γ)$ $(\alpha, \beta^2 - 4\gamma)$.

Pattern Recognition

Tangency conditions fix center parameters to match radius scale sizes instantly, reducing variable overhead in coordinate transformations.

Chapter Mix

Class 11 Maths: Circles

More Circles Previous-Year Questions — Page 4

Q4 jee_main_2024_31_jan_evening Equation of Tangent and Normal

Let a variable line passing through the centre of the circle

x² + y² - 16x - 4y = 0

, meet the positive co-ordinate axes at the point

A

and

B

. Then the minimum value of

OA + OB

, where

O

is the origin, is equal to

A. $12$ $12$
B. $18$ $18$
C. $20$ $20$
D. $24$ $24$

Solution

Related Formula

Intercept form of line: (x)/(a) + (y)/(b) = 1

Core Logic

Circle $x² + y² - 16x - 4y = 0$ $x^2 + y^2 - 16x - 4y = 0$ has its centre at $(8, 2)$ $(8, 2)$. Let the line passing through $(8, 2)$ $(8, 2)$ have slope $m$ $m$. Its equation is:

y - 2 = m(x - 8)

x-intercept ( $A$ $A$): set $y=0 -2 = m(x-8) x = 8 - (2)/(m)$ . y-intercept ( $B$ $B$): set $x=0 y = 2 - 8m$ $x=0 \implies y = 2 - 8m$. Sum of intercepts $OA + OB = (8 - (2)/(m)) + (2 - 8m) = 10 - (2)/(m) - 8m$ . To minimize, let $f(m) = 10 - (2)/(m) - 8m$ $f(m) = 10 - \frac{2}{m} - 8m$.

f'(m) = (2)/(m²) - 8 = 0 m² = (1)/(4)

Since the line meets the positive coordinate axes, intercepts must be positive, which requires $m < 0$ $m < 0$. Thus $m = -1/2$ $m = -1/2$. Substitute $m = -1/2$ $m = -1/2$:

OA + OB = 10 - (2)/(-1/2) - 8(-1/2) = 10 + 4 + 4 = 18

Pattern Recognition

AM-GM can also be applied: $8a + 2b = ab 1 = (8)/(a) + (2)/(b)$ . To minimize $a+b$ $a+b$, use Cauchy-Schwarz or standard differentiation. Differentiation directly yields intercept minima.

Chapter Mix

Class 11 Maths: Circles

Q4 jee_main_2024_31_jan_morning Intersection and Common Chords

If one of the diameters of the circle

x² + y² - 10x + 4y + 13 = 0

is a chord of another circle C, whose center is the point of intersection of the lines

2x + 3y = 12

and

3x - 2y = 5

, then the radius of the circle C is

A. $\sqrt(20)$ $\sqrt{20}$
B. $4$ $4$
C. $6$ $6$
D. $3\sqrt(2)$ $3\sqrt{2}$

Solution

Core Logic

Find the center of circle C by solving $2x + 3y = 12$ $2x + 3y = 12$ and $3x - 2y = 5$ $3x - 2y = 5$. Multiplying and subtracting yields $13x = 39 x = 3, y = 2$ $13x = 39 \implies x = 3, y = 2$. Center of C is $(3, 2)$ $(3, 2)$.

Step 1: Properties of Given Circle

Given circle: $x² + y² - 10x + 4y + 13 = 0$ $x^2 + y^2 - 10x + 4y + 13 = 0$. Center $M(5, -2)$ $M(5, -2)$. Radius $r = \sqrt(25 + 4 - 13) = 4$ $r = \sqrt{25 + 4 - 13} = 4$.

Intersection and Common Chords diagram for Q4 - JEE Main 2024 Morning

Step 2: Radius Calculation

The diameter of the first circle is a chord of circle C. Therefore, the distance between the two centers forms a right-angled triangle with the radius of C ( $CP$ $CP$) and the radius of the first circle ( $r = 4$ $r = 4$). Distance $CM = \sqrt((5-3)² + (-2-2)²) = \sqrt(4 + 16) = \sqrt(20)$ . Radius of circle C is $CP = \sqrt(CM² + r²) = \sqrt(20 + 16) = \sqrt(36) = 6$ .

Pattern Recognition

When a diameter of circle 1 is a chord of circle 2, the triangle formed by the centers and the point of intersection is a right-angled triangle at the center of circle 1.

Chapter Mix

Class 11 Maths: Circles Class 11 Maths: Straight Lines

Let the equation of the circle, which touches x-axis at the point (a, 0), $a > 0$ $a > 0$ and cuts off an intercept of length b on y-axis be $x² + y² - α x + β y + γ = 0$ . If the circle lies below x-axis, then the ordered pair $(2a, b²)$ $(2a, b^2)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Translating to General Equation Form

Step 2: Evaluating the Target Mapped Ordered Pair

Pattern Recognition

Chapter Mix

More Circles Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Properties of Given Circle

Step 2: Radius Calculation

Pattern Recognition

Chapter Mix

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Solution & Explanation

Related Formula

Core Logic

Step 1: Translating to General Equation Form

Step 2: Evaluating the Target Mapped Ordered Pair

Pattern Recognition

Chapter Mix

More Circles Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Properties of Given Circle

Step 2: Radius Calculation

Pattern Recognition

Chapter Mix

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