Let circle C$C$ be the image of x^2 + y^2 - 2x + 4y - 4 = 0$x^{2} + y^{2} - 2x + 4y - 4 = 0$ in the line 2x - 3y + 5 = 0$2x - 3y + 5 = 0$ and A$A$ be the point on C$C$ such that OA$OA$ is \parallel to the x-axis and A$A$ lies on the \right hand side of the centre O$O$ of C$C$. If B(alpha,beta)$B(\alpha,\beta)$, with beta < 4$\beta < 4$, lies on C$C$ such that the length of the arc AB$AB$ is (1/6)^textth$(1/6)^{\text{th}}$ of the perimeter of C$C$, then beta - sqrt3alpha$\beta - \sqrt{3}\alpha$ is equal to :
A.3$3$
B.3 + sqrt3$3 + \sqrt{3}$
C.4 - sqrt3$4 - \sqrt{3}$
D.4$4$
Solution & Explanation
### Related Formula
The coordinates for the reflection image of a point (x_1, y_1)$(x_1, y_1)$ across a standard line ax + by + c = 0$ax + by + c = 0$ are determined using:
fracx - x_1a = fracy - y_1b = frac-2(ax_1 + by_1 + c)a^2 + b^2$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$
### Core Logic
Find the center and radius of the original given circle:
x^2 + y^2 - 2x + 4y - 4 = 0 implies textCenter = (1, -2), \, r = sqrt1^2 + (-2)^2 - (-4) = 3$x^2 + y^2 - 2x + 4y - 4 = 0 \implies \text{Center} = (1, -2), \, r = \sqrt{1^2 + (-2)^2 - (-4)} = 3$Transformation and Reflection of Circles
Reflect the center point (1, -2)$(1, -2)$ across the line mirror 2x - 3y + 5 = 0$2x - 3y + 5 = 0$:
fracx - 12 = fracy + 2-3 = frac-2(2(1) - 3(-2) + 5)2^2 + (-3)^2 = frac-2(2 + 6 + 5)13 = -2$\frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{-2(2(1) - 3(-2) + 5)}{2^2 + (-3)^2} = \frac{-2(2 + 6 + 5)}{13} = -2$x - 1 = -4 implies x = -3$x - 1 = -4 \implies x = -3$y + 2 = 6 implies y = 4$y + 2 = 6 \implies y = 4$
Thus, the center O$O$ of the reflected circle C$C$ is (-3, 4)$(-3, 4)$, and its radius is preserved at r = 3$r = 3$.
### Step 1: Locate Point A
We are given that OA$OA$ is \parallel to the x-axis, meaning its y-coordinate matches the center. Since A$A$ lies to the \right of the center O(-3, 4)$O(-3, 4)$:
A = (-3 + r, \, 4) = (-3 + 3, \, 4) = (0, 4)$A = (-3 + r, \, 4) = (-3 + 3, \, 4) = (0, 4)$
### Step 2: Determine Angular Position of Point B
The arc length AB$AB$ is given as frac16$\frac{1}{6}$ of the total perimeter:
textArc length = rtheta = frac16(2pi r) implies theta = fracpi3 = 60^circ$\text{Arc length} = r\theta = \frac{1}{6}(2\pi r) \implies \theta = \frac{\pi}{3} = 60^\circ$Transformation and Reflection of Circles
Using parametric coordinates relative to center O(-3, 4)$O(-3, 4)$ with radius r = 3$r = 3$:
alpha = -3 + 3costheta, quad beta = 4 + 3sintheta$\alpha = -3 + 3\cos\theta, \quad \beta = 4 + 3\sin\theta$
Since beta < 4$\beta < 4$, the \angle theta$\theta$ must point downwards into the negative quadrant relative to A$A$, meaning theta = -60^circ = -fracpi3$\theta = -60^\circ = -\frac{\pi}{3}$:
alpha = -3 + 3cosleft(-fracpi3right) = -3 + 3left(frac12
ight) = -frac32$\alpha = -3 + 3\cos\left(-\frac{\pi}{3}\right) = -3 + 3\left(\frac{1}{2}
ight) = -\frac{3}{2}$beta = 4 + 3sinleft(-fracpi3right) = 4 - frac3sqrt32$\beta = 4 + 3\sin\left(-\frac{\pi}{3}\right) = 4 - \frac{3\sqrt{3}}{2}$
### Step 3: Evaluate Final Algebraic Value
Substitute the determined coordinates into the target expression:
beta - sqrt3alpha = left(4 - frac3sqrt32right) - sqrt3left(-frac32
ight)$\beta - \sqrt{3}\alpha = \left(4 - \frac{3\sqrt{3}}{2}\right) - \sqrt{3}\left(-\frac{3}{2}
ight)$beta - sqrt3alpha = 4 - frac3sqrt32 + frac3sqrt32 = 4$\beta - \sqrt{3}\alpha = 4 - \frac{3\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 4$
### Pattern Recognition
Whenever parametric configurations on a circle involve radical coordinate multipliers like beta - sqrt3alpha$\beta - \sqrt{3}\alpha$, using angular vectors centered at the origin of the circle avoids setting up and solving long distance equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Circles
Keywords:#circle reflection line image#parametric circle coordinates#JEE Main 2025 Morning Q65#arc length angular sector
More Circles Previous-Year Questions
Q752025Tangent Properties of Circles
The absolute difference between the squares of the radii of the two circles passing through the point (-9, 4)$(-9, 4)$ and touching the lines x + y = 3$x + y = 3$ and x - y = 3$x - y = 3$, is equal to ________.
Numerical Answer.Answer: 768 to 768
Solution
### Related Formula
Perpendicular distance from point (x_0, y_0)$(x_0, y_0)$ to line Ax + By + C = 0$Ax + By + C = 0$:
d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$
### Core Logic
Since the circle touches two symmetric intersecting lines, its center must lie on their angle bisector (x$x$-axis). Use this property to find the center parameters. Tangent Properties of Circles diagram for Q75 - JEE Main 2025 Morning
### Step 1: Establish Center and Radius Equations
The lines are x+y-3=0$x+y-3=0$ and x-y-3=0$x-y-3=0$. The intersection point is (3,0)$(3,0)$, and the bisector line is the x$x$-axis.
Let the center be C(a, 0)$C(a, 0)$. The radius r$r$ is the perpendicular distance to either line:
r = frac|a - 0 - 3|sqrt1^2 + 1^2 = frac|a - 3|sqrt2$r = \frac{|a - 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{|a - 3|}{\sqrt{2}}$
### Step 2: Apply Point Passage Constraint
The circle equation is (x - a)^2 + y^2 = r^2$(x - a)^2 + y^2 = r^2$. Substitute the given passage point (-9, 4)$(-9, 4)$:
(-9 - a)^2 + 4^2 = left(fraca - 3sqrt2right)^2$(-9 - a)^2 + 4^2 = \left(\frac{a - 3}{\sqrt{2}}\right)^2$2left(a^2 + 18a + 81 + 16right) = a^2 - 6a + 9$2\left(a^2 + 18a + 81 + 16\right) = a^2 - 6a + 9$2a^2 + 36a + 194 = a^2 - 6a + 9 implies a^2 + 42a + 185 = 0$2a^2 + 36a + 194 = a^2 - 6a + 9 \implies a^2 + 42a + 185 = 0$
### Step 3: Solve for Quadratic Roots
Factor the quadratic equation:
(a + 37)(a + 5) = 0 implies a_1 = -37, quad a_2 = -5$(a + 37)(a + 5) = 0 \implies a_1 = -37, \quad a_2 = -5$
### Step 4: Compute Radii Squares Difference
Find the corresponding radius value for each root:
r_1 = frac|-37 - 3|sqrt2 = frac40sqrt2 = 20sqrt2 implies r_1^2 = 800$r_1 = \frac{|-37 - 3|}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2} \implies r_1^2 = 800$r_2 = frac|-5 - 3|sqrt2 = frac8sqrt2 = 4sqrt2 implies r_2^2 = 32$r_2 = \frac{|-5 - 3|}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \implies r_2^2 = 32$
The absolute difference between their squares is:
left|r_1^2 - r_2^2right| = |800 - 32| = 768$\left|r_1^2 - r_2^2\right| = |800 - 32| = 768$
### Pattern Recognition
Recognizing that the center must lie on the line of symmetry (x$x$-axis) eliminates one variable parameter immediately, reducing a difficult geometric system to a simple single-variable quadratic equation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Circles
Class 11 Mathematics: Straight Lines
Q2025Tangent and Normal to a Circle
Let C_1$C_1$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let C_2$C_2$ be the circle with centre (1, 3) that touches C_1$C_1$ externally at the point (alpha, beta)$(\alpha, \beta)$. If (beta - alpha)^2 = fracmn$(\beta - \alpha)^2 = \frac{m}{n}$, gcd(m, n) = 1$\gcd(m, n) = 1$, then m + n$m + n$ is equal to:
A.9$9$
B.13$13$
C.22$22$
D.31$31$
Solution
### Related Formula
For a circle in the third quadrant touching both coordinate axes, the center layout is (-r, -r)$(-r, -r)$ and equation looks like:
(x + r)^2 + (y + r)^2 = r^2$(x + r)^2 + (y + r)^2 = r^2$
For external contact between circles C_1$C_1$ and C_2$C_2$, the distance between centers equals the sum of their radii:
C_1C_2 = r_1 + r_2$C_1C_2 = r_1 + r_2$
### Core Logic
Circle C_1$C_1$ has radius r_1 = 3$r_1 = 3$ and touches both axes in the third quadrant, so its center is A(-3, -3)$A(-3, -3)$.
Circle C_2$C_2$ has center B(1, 3)$B(1, 3)$.
The distance between centers A$A$ and B$B$ is:
AB = sqrt(1 - (-3))^2 + (3 - (-3))^2 = sqrt4^2 + 6^2 = sqrt16 + 36 = sqrt52 = 2sqrt13$AB = \sqrt{(1 - (-3))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$
### Step 1: Determine Radius of Circle 2
Tangent and Normal to a Circle diagram for Q59 - JEE Main 2025 Morning
Since the circles touch externally:
AB = r_1 + r_2 implies 2sqrt13 = 3 + r_2 implies r_2 = 2sqrt13 - 3$AB = r_1 + r_2 \implies 2\sqrt{13} = 3 + r_2 \implies r_2 = 2\sqrt{13} - 3$
### Step 2: Locate the Contact Point via Section Formula
The point of contact P(alpha, beta)$P(\alpha, \beta)$ divides the line segment joining centers A(-3, -3)$A(-3, -3)$ and B(1, 3)$B(1, 3)$ internally in the ratio r_1 : r_2 = 3 : (2sqrt13 - 3)$r_1 : r_2 = 3 : (2\sqrt{13} - 3)$.
Using the internal section formula:
alpha = frac3(1) + (2sqrt13 - 3)(-3)3 + (2sqrt13 - 3) = frac3 - 6sqrt13 + 92sqrt13 = frac12 - 6sqrt13 + 02sqrt13 = frac6 - 3sqrt13sqrt13$\alpha = \frac{3(1) + (2\sqrt{13} - 3)(-3)}{3 + (2\sqrt{13} - 3)} = \frac{3 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{12 - 6\sqrt{13} + 0}{2\sqrt{13}} = \frac{6 - 3\sqrt{13}}{\sqrt{13}}$beta = frac3(3) + (2sqrt13 - 3)(-3)3 + (2sqrt13 - 3) = frac9 - 6sqrt13 + 92sqrt13 = frac18 - 6sqrt132sqrt13 = frac9 - 3sqrt13sqrt13$\beta = \frac{3(3) + (2\sqrt{13} - 3)(-3)}{3 + (2\sqrt{13} - 3)} = \frac{9 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{18 - 6\sqrt{13}}{2\sqrt{13}} = \frac{9 - 3\sqrt{13}}{\sqrt{13}}$
### Step 3: Calculate the Difference Value
Find (beta - alpha)^2$(\beta - \alpha)^2$:
beta - alpha = frac9 - 3sqrt13sqrt13 - frac6 - 3sqrt13sqrt13 = frac3sqrt13$\beta - \alpha = \frac{9 - 3\sqrt{13}}{\sqrt{13}} - \frac{6 - 3\sqrt{13}}{\sqrt{13}} = \frac{3}{\sqrt{13}}$(beta - alpha)^2 = left(frac3sqrt13right)^2 = frac913$(\beta - \alpha)^2 = \left(\frac{3}{\sqrt{13}}\right)^2 = \frac{9}{13}$
Comparing with fracmn$\frac{m}{n}$ where gcd(m, n) = 1$\gcd(m, n) = 1$ gives m = 9, n = 13$m = 9, n = 13$.
m + n = 9 + 13 = 22$m + n = 9 + 13 = 22$
### Pattern Recognition
Notice that computing (beta - alpha)$(\beta - \alpha)$ directly cancels out the irrational sqrt13$\sqrt{13}$ term from the numerator before squaring, saving a significant amount of tedious arithmetic expansion.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Coordinate Geometry
Class 11 Mathematics: Circles
Q642025Equation of a Circle
Let a circle C$C$ pass through the points (4,2)$(4,2)$ and (0,2)$(0,2)$, and its centre lie on 3x + 2y + 2 = 0$3x + 2y + 2 = 0$. Then the length of the chord, of the circle C$C$, whose mid-point is (1,2)$(1,2)$, is:
A.sqrt3$\sqrt{3}$
B.2sqrt3$2\sqrt{3}$
C.4sqrt2$4\sqrt{2}$
D.2sqrt2$2\sqrt{2}$
Solution
### Related Formula
Length of a chord with perpendicular distance d$d$ from the center of a circle of radius r$r$ is:
textLength = 2sqrtr^2 - d^2$\text{Length} = 2\sqrt{r^2 - d^2}$
### Core Logic
Points A(4,2)$A(4,2)$ and B(0,2)$B(0,2)$ have the same y-coordinate, meaning chord AB$AB$ is horizontal. The perpendicular bisector of a horizontal chord is vertical.
Equation of a Circle diagram for Q64 - JEE Main 2025 Evening
Midpoint of AB$AB$ is M(2,2)$M(2,2)$. Thus, the vertical line passing through the center is x = 2$x = 2$.
### Step 1: Identify Center and Radius
Since the center lies on the line 3x + 2y + 2 = 0$3x + 2y + 2 = 0$, substitute x = 2$x = 2$ to find the y-coordinate:
3(2) + 2y + 2 = 0 implies 2y = -8 implies y = -4$3(2) + 2y + 2 = 0 \implies 2y = -8 \implies y = -4$
So, Center O = (2, -4)$O = (2, -4)$.
Calculate radius r$r$ using point B(0,2)$B(0,2)$:
r = OB = sqrt(2 - 0)^2 + (-4 - 2)^2 = sqrt4 + 36 = sqrt40$r = OB = \sqrt{(2 - 0)^2 + (-4 - 2)^2} = \sqrt{4 + 36} = \sqrt{40}$
### Step 2: Find Target Chord Length
The targeted chord has a given midpoint N(1,2)$N(1,2)$. Distance from center O(2,-4)$O(2,-4)$ to N(1,2)$N(1,2)$:
d = ON = sqrt(2 - 1)^2 + (-4 - 2)^2 = sqrt1 + 36 = sqrt37$d = ON = \sqrt{(2 - 1)^2 + (-4 - 2)^2} = \sqrt{1 + 36} = \sqrt{37}$textLength of chord = 2sqrtr^2 - d^2 = 2sqrt40 - 37 = 2sqrt3$\text{Length of chord} = 2\sqrt{r^2 - d^2} = 2\sqrt{40 - 37} = 2\sqrt{3}$
### Pattern Recognition
Points sharing a coordinate define standard vertical or horizontal perpendicular configurations immediately. Always exploit geometrical configurations before jumping into standard circle equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Circles
Q582025Circles Touching Axes and Intercepts
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 x^2 + y^2 - alpha x + beta y + gamma = 0$x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below x-axis, then the ordered pair (2a, b^2)$(2a, b^2)$ is equal to:
(1) (alpha, beta^2 + 4gamma)$(alpha, \beta^2 + 4gamma)$
(2) (gamma, beta^2 - 4alpha)$(gamma, \beta^2 - 4alpha)$
(3) (gamma, beta^2 + 4alpha)$(gamma, \beta^2 + 4alpha)$
(4) (alpha, beta^2 - 4gamma)$(alpha, \beta^2 - 4gamma)$
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