Exams › JEE Main › Maths

Find the sum of all possible slopes of tangent lines to the circle (x - 2)² + (y - 3)² = 1 that pass through the origin.

1. 4
2. 3
3. 12/5
4. 6

Correct answer: 4

Solution

Line through origin: y = mx, or mx - y = 0. Distance from center (2,3) to this line = |2m-3|/sqrt(m²+1) = 1. Squaring: (2m-3)² = m²+1 => 4m² - 12m + 9 = m² + 1 => 3m² - 12m + 8 = 0. By Vieta's formulas, sum of roots = 12/3 = 4.

Related JEE Main Maths questions

• If the line 3x + 4y + c = 0 is tangent to the circle x² + y² - 2x - 2y - 2 = 0, what is the maximum possible value of c?
• The circle S: x² + y² + k*x + (k+1)*y - (k+1) = 0 passes through two fixed points for all real values of k. If the minimum radius of circle S is 1/sqrt(P), find the value of P.
• Consider the two circles C1: x² + y² - 4x + 6y + 11 = 0 and C2: x² + y² - 2x + 8y + 13 = 0. A circle passes through both points of intersection of C1 and C2, and also passes through the origin. Find the x-coordinate of the centre of this circle.
• Find all circles that pass through the point (3, -6) and are tangent to both coordinate axes.
• Let AB be a variable chord of length 5 to the circle x² + y² = 25/2. A triangle ABC is formed such that BC = 4 and CA = 3. If the locus of vertex C is x² + y² = a, then the only possible integral value of a is
• Tangents are drawn to the circle x² + y² = 1 at its intersection points (distinct) with the circle x² + y² + (lambda - 3)x + (2*lambda + 2)y + 2 = 0. As lambda varies, the locus of the intersection point of the pair of tangents is a straight line. Find the slope of that straight line.

⚔️ Practice JEE Main Maths free + battle 1v1 →