Exams › JEE Advanced › Maths

A square of side 3/sqrt(10) is inscribed in a quarter circle of radius R such that two adjacent vertices lie on the two straight edges (radii) of the quarter circle and the other two vertices lie on the arc. Find the value of 2R².

1. 3.5
2. 4.5
3. 6.25
4. 5

Correct answer: 4.5

Solution

With A=(a,0), B=(0,a) on the axes and side s = a*sqrt(2) = 3/sqrt(10), we get a = 3/sqrt(20). The other two vertices are D=(2a, a) and C=(a, 2a). They lie on the arc: R² = (2a)² + a² = 5a² = 5*(9/20) = 9/4. So 2R² = 9/2 = 4.5.

Related JEE Advanced 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 Advanced Maths free + battle 1v1 →