Let PQ and MN be two straight lines touching the circle 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 . Then the locus of the point of intersection of the lines PQ and MN is:

Solution & Explanation

### Related Formula For external tangents from point R forming angle 2theta at the center, the distance d from center to intersection point obeys cos theta = fracrd. ### Core Logic Given circle: x^2 + y^2 - 4x - 6y - 3 = 0 Center O = (2, 3) Radius r = sqrt(-2)^2 + (-3)^2 - (-3) = sqrt4 + 9 + 3 = sqrt16 = 4. The tangents PQ and MN intersect at some point R(h, k). The radius vectors OA and OB subtend angle angle AOB = fracpi3 = 60^circ at the center. The line joining O and R bisects the angle angle AOB. Thus, angle AOR = 30^circ. ### Step 1: Apply Trigonometric Relations
Circle tangents intersection locus diagram for Q15 - JEE Main 2026 Morning
Circle tangents intersection locus diagram for Q15 - JEE Main 2026 Morning
In the right-angled triangle Delta AOR, OA is the radius (r = 4) and OR is the hypotenuse. cos(30^circ) = fracOAOR = fracrOR fracsqrt32 = frac4OR Rightarrow OR = frac8sqrt3 ### Step 2: Construct the Locus Equation The distance squared between O(2,3) and R(h,k) is OR^2: OR^2 = (h - 2)^2 + (k - 3)^2 = left(frac8sqrt3right)^2 (h - 2)^2 + (k - 3)^2 = frac643 h^2 - 4h + 4 + k^2 - 6k + 9 = frac643 3(h^2 + k^2 - 4h - 6k + 13) = 64 3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0 3(h^2 + k^2) - 12h - 18k - 25 = 0 ### Step 3: Generalize the Equation Replace (h, k) with (x, y) for the general locus: 3(x^2 + y^2) - 12x - 18y - 25 = 0 ### Pattern Recognition The locus of the intersection of tangents enclosing a constant angle is simply a concentric circle. Its radius expands by 1/sin(alpha/2) or 1/cos(theta) depending on whether the angle is measured at intersection or center. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

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Circles

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Q4 jee_main_2024_31_jan_evening Equation of Tangent and Normal
Let a variable line passing through the centre of the circle x^2 + y^2 - 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
  • B. 18
  • C. 20
  • D. 24

Solution

### Related Formula textIntercept form of line: fracxa + fracyb = 1 ### Core Logic Circle x^2 + y^2 - 16x - 4y = 0 has its centre at (8, 2). Let the line passing through (8, 2) have slope m. Its equation is: y - 2 = m(x - 8) x-intercept (A): set y=0 implies -2 = m(x-8) implies x = 8 - frac2m. y-intercept (B): set x=0 implies y = 2 - 8m. Sum of intercepts OA + OB = (8 - frac2m) + (2 - 8m) = 10 - frac2m - 8m. To minimize, let f(m) = 10 - frac2m - 8m. f'(m) = frac2m^2 - 8 = 0 implies m^2 = frac14 Since the line meets the positive coordinate axes, intercepts must be positive, which requires m < 0. Thus m = -1/2. Substitute m = -1/2: OA + OB = 10 - frac2-1/2 - 8(-1/2) = 10 + 4 + 4 = 18 ### Pattern Recognition AM-GM can also be applied: 8a + 2b = ab implies 1 = frac8a + frac2b. To minimize a+b, use Cauchy-Schwarz or standard differentiation. Differentiation directly yields intercept minima. ### Evaluation Rubric / Model Answer null ### 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 x^2 + y^2 - 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. sqrt20
  • B. 4
  • C. 6
  • D. 3sqrt2

Solution

### Core Logic Find the center of circle C by solving 2x + 3y = 12 and 3x - 2y = 5. Multiplying and subtracting yields 13x = 39 implies x = 3, y = 2. Center of C is (3, 2). ### Step 1: Properties of Given Circle Given circle: x^2 + y^2 - 10x + 4y + 13 = 0. Center M(5, -2). Radius r = sqrt25 + 4 - 13 = 4.
Intersection and Common Chords diagram for Q4 - JEE Main 2024 Morning
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) and the radius of the first circle (r = 4). Distance CM = sqrt(5-3)^2 + (-2-2)^2 = sqrt4 + 16 = sqrt20. Radius of circle C is CP = sqrtCM^2 + r^2 = sqrt20 + 16 = sqrt36 = 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles Class 11 Maths: Straight Lines

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