Exams › JEE Main › Maths

Find the equation of the circle that touches the line x + y = 5 at the point N(-2, 7) and cuts the circle x² + y² + 4x - 6y + 9 = 0 orthogonally.

1. x² + y² + 7x - 11y + 38 = 0
2. x² + y² = 53
3. x² + y² + x - y - 44 = 0
4. x² + y² - x + y - 62 = 0

Correct answer: x² + y² + 7x - 11y + 38 = 0

Solution

Writing the tangent family (x+2)²+(y-7)²+lambda(x+y-5)=0 and applying the orthogonality condition with the given circle yields lambda = 3, giving x² + y² + 7x - 11y + 38 = 0.

Related JEE Main Maths questions

• For the pair of parallel straight lines represented by 9x² - 6xy + y² + 18x - 6y + 8 = 0, what is the separation between them?
• An ellipse has its two foci 10 units apart, and the length of its latus rectum is 15. If its axes are taken as the coordinate axes, which equation represents the ellipse?
• On the segment joining A(0, 0) and B(3a, 0), choose points P and Q so that AP = PQ = QB. Three circles are then constructed with AP, PQ, and QB as their respective diameters. If a point S is such that the sum of the squares of the tangents drawn from S to these three circles is b², then the locus of S is
• Find the coordinates of the midpoint of the chord cut by the circle x² + y² + 4x - 2y - 3 = 0 on the line y = x + 2.
• A hyperbola has a transverse axis of length 2 sin θ and is confocal with the ellipse 3x² + 4y² = 12. Its equation is
• Three points E, F and G are chosen on the parabola y² = 4ax such that their y-coordinates form a geometric progression. The point where the tangents at E and G meet lies on the

⚔️ Practice JEE Main Maths free + battle 1v1 →