The problem involves three concentric circles with radii in an arithmetic progression (AP).
Key Information:In an arithmetic progression, the difference between consecutive terms is constant. For three terms $a, b, c$, they are in AP if $b - a = c - b$. This simplifies to the property that the middle term ($b$) is the average of the first ($a$) and the third ($c$) term:
$ b = \frac{a + c}{2} $
Applying this property to the radii of the concentric circles ($r_1, r_2, r_3$):
The middle radius ($r_2$) is the average of the innermost radius ($r_1$) and the outermost radius ($r_3$).
$ r_2 = \frac{r_1 + r_3}{2} $
Substitute the given values:
$ r_2 = \frac{7 \text{ cm} + 21 \text{ cm}}{2} $
$ r_2 = \frac{28 \text{ cm}}{2} $
$ r_2 = 14 \text{ cm} $
The radius of the middle circle is 14 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.