Let PQ and MN be two straight lines touching the circle x^2 + y^2 - 4x - 6y - 3 = 0$x^{2} + y^{2} - 4x - 6y - 3 = 0$ at the points A and B respectively. Let O be the centre of the circle and angle AOB = pi/3$\angle AOB = \pi/3$ . Then the locus of the point of intersection of the lines PQ and MN is:
Keywords:#locus of the point of intersection#JEE Main 2026 Morning Q15#Circles JEE Main 2026#Locus of Intersection of Tangents JEE Main 2026
More Circles Previous-Year Questions
Q75jee_main_2025_02_april_morningTangent 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
Qjee_main_2025_07_april_morningTangent 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
Q64jee_main_2025_29_jan_eveningEquation 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
Q58jee_main_2025_28_jan_morningCircles 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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