This problem involves finding the length of a chord within a circle given its radius and its distance from the center. We can visualize a right-angled triangle formed by:
Let the radius of the circle be r = 15 cm.
Let the distance of the chord from the center be d = 9 cm.
Let half the length of the chord be x. The full length of the chord will be 2x.
According to the Pythagorean theorem in the right-angled triangle:
$r^2 = d^2 + x^2$
Substitute the given values:
$15^2 = 9^2 + x^2$
$225 = 81 + x^2$
Solve for $x^2$:
$x^2 = 225 - 81$
$x^2 = 144$
Find the value of x:
$x = \sqrt{144}$
$x = 12 \text{ cm}$
The length of the chord is twice the value of x:
Chord Length = $2x$
Chord Length = $2 \times 12$ cm
Chord Length = 24 cm
The length of the chord is 24 cm.
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.