JEE Main Mathematics: Equations of tangent, normal and chord of a circle Practice Questions

Correct answer: C — The chord is perpendicular to the diameter.

If a chord is tangent to the semicircle at its midpoint, the tangent line (the chord) must be perpendicular to the radius (which is part of the diameter) at that point. Therefore, the chord is perpendicular to the diameter.

Correct answer: C — Equal

When two chords are parallel in a circle, the arcs intercepted between them are equal (i.e., arc AC = arc BD).

Correct answer: A — The product is -1

A radius drawn to the point of contact is perpendicular to the tangent at that point. If two lines are perpendicular, the product of their slopes is -1.

Correct answer: D — ax + by = r²

The equation of the chord of contact from an external point (x₁, y₁) to the circle x² + y² = r² is x x₁ + y y₁ = r². Thus, for point (a, b), the equation is ax + by = r².

Correct answer: C — The lengths of the tangent segments

A fundamental theorem in circle geometry states that the lengths of the two tangents drawn from an external point to a circle are equal.

Correct answer: C — 2

Let R be the radius (10 cm) and L be the chord length (16 cm). The distance (d) from the center to the chord is found using the Pythagorean theorem on the right triangle formed by the radius, half the chord, and the distance d. Half the chord is 16/2 = 8 cm. d² + 8² = 10². d² = 100 - 64 = 36. Thus, d=6 cm.

Correct answer: B — 90°

The angle subtended by a diameter at any point on the circumference is always 90°. This is a fundamental theorem related to circles (Thales' Theorem).

Correct answer: D — (2,0)

The standard equation of a circle is (x-h)² + (y-k)² = r², where (h, k) is the center. Comparing with (x-2)² + (y-0)² = 25, the center is (h, k) = (2, 0).

Correct answer: D — The midpoint of the chord

The perpendicular drawn from the center of a circle to a chord always bisects the chord. Therefore, the foot of the perpendicular from the center to the chord is the midpoint of that chord.

Correct answer: C — 3 units

Let O be the center, P be the point of contact, and the tangent length be PT. Since the radius OP is perpendicular to the tangent PT, triangle OPT is a right-angled triangle. OP² + PT² = OP² + 12². This setup requires the distance from the center, which is not directly given by this setup unless we consider the chord property. Reinterpreting the constraint: If the problem intends a direct application of Pythagoras on the radius/tangent relationship, the distance from P to the center O is simply the radius. If the setup is: Tangent length = 12, Radius = 5. The point P is ON the circle, so the distance from P to O is the radius. (This tests basic Pythagorean concept related to radius and tangent.)