Circles — Chapter Test | Class 11 IMO

Q1

What is the point of contact of the tangent 3x + 4y = 25 to the circle x² + y² = 25?

The tangent at (x1, y1) is xx1 + yy1 = 25. Comparing with 3x + 4y = 25, we get x1 = 3 and y1 = 4. The point is (3, 4).

Q2

The locus of the point of intersection of perpendicular tangents to a circle is called its:

By definition, the locus of the point of intersection of two perpendicular tangents to a given conic (including a circle) is its director circle.

Q3

The line y=2 is tangent to which circle?

Circle centre=(0,2), radius=2. Distance to y=2 is 0, not tangent. For x²+y²−4y=0, tangent at top point y=4? Actually line y=2 passes through centre. Hence not tangent. For x²+y²−2y=0, centre=(0,1), radius=1 and distance to y=2 is 1, tangent.

Q4

A farmer in Punjab has a circular wheat field represented by x² + y² = 400 (in metres). A straight irrigation canal runs through the field along the line x = 12. Find the length of the canal inside the field.

Substitute x = 12 into the field's equation: 144 + y² = 400 → y² = 256 → y = ±16. The coordinates where the canal enters and exits are (12, 16) and (12, −16). Length = 16 − (−16) = 32 m.

Q5

Find the equations of the tangents to the circle x² + y² = 25 that are parallel to the line 3x − 4y = 0.

The given line has slope m = 3/4. Tangents parallel to it have equations y = (3/4)x ± a√(1 + m²). Here a = 5. So, y = 3/4 x ± 5√(1 + 9/16) = 3/4 x ± 5(5/4). Multiplying by 4 gives 4y = 3x ± 25, or 3x − 4y = ± 25.

Q6

What is the radius of the circle (x + 1)² + (y − 2)² = 49?

Comparing with (x − h)² + (y − k)² = r², we get r² = 49. Taking the positive square root, the radius r = 7.

Q7

If the tangent at a point P on the circle x² + y² + 6x + 6y = 2 meets the straight line 5x − 2y + 6 = 0 at a point Q on the y-axis, then the length of PQ is:

Q lies on the y-axis, so x = 0. Substituting in the line equation gives −2y + 6 = 0 → y = 3. So, Q is (0, 3). PQ is the length of the tangent from Q to the circle: √S1 = √(0² + 3² + 6(0) + 6(3) − 2) = √(9 + 18 − 2) = √25 = 5.

Q8

Find the centre of the circle passing through the points (1, 1), (2, 2), and (3, 3).

The points (1, 1), (2, 2), and (3, 3) are collinear (they lie on the line y = x). A circle cannot pass through three collinear points. Thus, such a circle does not exist.

Q9

The angle subtended by a semicircle at any point on its circumference is always:

By Thales's theorem, the angle inscribed in a semicircle is always a right angle, which is 90°.

Q10

What is the area of an equilateral triangle inscribed in the circle x² + y² = a²?

The circumradius of the triangle is R = a. For an equilateral triangle, side length s = R√3 = a√3. Area = (√3/4)s² = (√3/4)(3a²) = (3√3/4)a².

Q11

The correct radius of the park is:

Radius²=25, therefore radius=5.

Q12

What is the maximum distance from the point (10, 7) to the circle x² + y² − 4x − 2y − 20 = 0?

Centre is (2, 1) and r = √(4 + 1 + 20) = 5. Distance of point to centre d = √((10 − 2)² + (7 − 1)²) = √(64 + 36) = 10. Max distance = d + r = 10 + 5 = 15.

Q13

A radar station at Mumbai airport is at the origin and tracks flights up to 50 km. A flight follows the linear path 3x + 4y = 150. For what distance is the flight within the radar's tracking range?

The radar's range is a circle x² + y² = 50². The distance of the flight path from the origin is d = |−150|/√(3² + 4²) = 150/5 = 30. The length of the chord (tracked path) is 2√(r² − d²) = 2√(50² − 30²) = 2×40 = 80 km.

Q14

Find the equation of the circle with centre (2, −3) and radius 4.

Equation is (x − 2)² + (y + 3)² = 16. Expanding gives x² − 4x + 4 + y² + 6y + 9 = 16, which simplifies to x² + y² − 4x + 6y − 3 = 0.

Q15

What is the equation of the polar of the point (1, 2) with respect to the circle x² + y² = 4?

The polar of (x1, y1) with respect to x² + y² = a² is xx1 + yy1 = a². Substitute x1 = 1, y1 = 2, and a² = 4 to get x(1) + y(2) = 4, or x + 2y = 4.