The relationship between the angle subtended by a chord at the center of a circle and the angle subtended at the circumference is defined by a key circle theorem.
The angle subtended by an arc or a chord at the center of the circle is twice the angle subtended by the same arc or chord at any point on the circumference. Importantly, this applies when the points are on the same side of the chord relative to the center.
Mathematically, if $\theta_c$ is the angle at the center and $\theta_a$ is the angle at the circumference:
$ \theta_c = 2 \times \theta_a $Or conversely:
$ \theta_a = \frac{\theta_c}{2} $Given:
We need to find the angle subtended at the circumference on the same side ($\theta_a$).
Using the theorem:
$ \theta_a = \frac{\theta_c}{2} $ $ \theta_a = \frac{75^{\circ}}{2} $ $ \theta_a = 37.5^{\circ} $Therefore, the angle subtended by the chord at the circumference on the same side is $37.5^{\circ}$.
If a tangent to a circle from a point P meets the circle at A with AP = 15 cm. Given that the radius of the circle is 8 cm, find the distance of P from the centre of the circle.
In a circle with a radius of 10 cm. XY and PQ are two parallel chords 12 cm and 16 cm in length, respectively. The two chords are situated on the opposite sides of the centre. The distance between the chords is:
Find the equation of the tangents to the circle x2 + y2 = 9 at x = 2.
If a chord of length 24 cm is at a distance of 5 cm from centre, then find the radius of the circle.
Let \( C_1 \) and \( C_2 \) be two circles which do not externally touch and intersect each other and \( O_1 \), and \( O_2 \) be the centers of the circles, respectively. Let AB be the common transverse tangent to the circles such that P, Q are the points of tangency respectively to \( C_1 \), \( C_2 \). Let R be the point of intersection of \( O_1 O_2 \) and AB. If \( \angle PO_1R = 60^\circ \), find \( \angle QO_2R \) and \( \angle QRO_2 \) respectively.