The absolute difference between the squares of the radii of the two circles passing through the point (-9, 4) and touching the lines x + y = 3 and x - y = 3, is equal to ________.

Numerical Answer Type:
Enter a numerical value Answer: 768 to 768 +4 marks

Solution & Explanation

### Related Formula Perpendicular distance from point (x_0, y_0) to line Ax + By + C = 0: d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2 ### Core Logic Since the circle touches two symmetric intersecting lines, its center must lie on their angle bisector (x-axis). Use this property to find the center parameters.
Tangent Properties of Circles diagram for Q75 - JEE Main 2025 Morning
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 and x-y-3=0. The intersection point is (3,0), and the bisector line is the x-axis. Let the center be C(a, 0). The radius r is the perpendicular distance to either line: r = frac|a - 0 - 3|sqrt1^2 + 1^2 = frac|a - 3|sqrt2 ### Step 2: Apply Point Passage Constraint The circle equation is (x - a)^2 + y^2 = r^2. Substitute the given passage point (-9, 4): (-9 - a)^2 + 4^2 = left(fraca - 3sqrt2right)^2 2left(a^2 + 18a + 81 + 16right) = a^2 - 6a + 9 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 ### 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_2 = frac|-5 - 3|sqrt2 = frac8sqrt2 = 4sqrt2 implies r_2^2 = 32 The absolute difference between their squares is: left|r_1^2 - r_2^2right| = |800 - 32| = 768 ### Pattern Recognition Recognizing that the center must lie on the line of symmetry (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

Reference Study Guides

More Circles Previous-Year Questions — Page 3

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

More Circles Questions — jee_main_2025_02_april_morning

Practice all Circles previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)