Equal Chords and their Distances from the Centre

In a circle with radius $13$$\text{ cm}$, two equal chords are at a distance of $5$$\text{ cm}$ from the centre. If the lengths of the chords is $k \mathrm{cm}$, then find the value of $k$.

Find the ratio between the chords which are equidistant from the centre of a circle.

In given figure, AB and CD are equal chords of a circle with centre O. If AB and CD meet at E (outside the circle), prove that BE = DE.

$PQR$ is an equilateral triangle inscribed in a circle. $A$ and $D$ are mid points of arcs $PQ$ and $PR$ respectively. Prove that $AB=BC=CD$.

A straight line is drawn cutting two circles of equal radii and passing through the mid point M of the line joining their centres O and O'. Prove that the chords PQ and RS, which are intercepted by the two circles, are equal.

PQ and RS are two equal chords of a circle with centre O. If PQ and RS, on being produced meet at T outside the circle, prove that TQ = TS

PQ and RS are two equal chords of a circle with centre O. If PQ and RS, on being produced meet at T outside the circle, prove that PT = RT

Two equal chords $AB$ and $CD$ of a circle $C\left(O,r\right)$ intersect at a point $P$ within a circle then $\angle OPL=$

Two equal chords $AB$ and $CD$ of a circle $C\left(O,r\right)$ intersect at a point $P$ within a circle then $DP=$

Two equal chords $AB$ and $CD$ of a circle $C\left(O,r\right)$ intersect at a point $P$ within a circle then $AP=$

Three chords $AB,CD$ and $EF$ of a circle are respectively $3\mathrm{cm},3.5\mathrm{cm}$ and $3.8\mathrm{cm}$ away from the center. Which of the following is correct?

Which of the following statements is true for the longest chord of a circle?

Prove that, of any two chords of a circle, the greater chord is nearer to the centre.
 

Circles with centres $\mathrm{P}$ and $\mathrm{Q}$ intersect at points $\mathrm{A}$ and $\mathrm{B}$ as shown in the figure. $\mathrm{CBD}$ is a line segment and $\mathrm{EBM}$ is tangent to the circle, with centre $\mathrm{Q},$ at point $\mathrm{B}.$ If the circles are congruent, show that $\mathrm{CE}=\mathrm{BD}.$

Two congruent circles of centres $\mathrm{O}$ and $\mathrm{O}\text{'}$ intersects each other at point $\mathrm{A}$ and $\mathrm{A}\text{'}$, then prove that $\angle \mathrm{AOB}=\angle \mathrm{AO}\text{'}\mathrm{B}$.

If two chords $\mathrm{AB}$ and $\mathrm{CD}$ are $4\mathrm{cm}$ away from the centre of a circle, then $\mathrm{AB}=\mathrm{CD}$.

The chords of a circle of length $10 \mathrm{cm}$ and $8 \mathrm{cm}$. Which apart from the centre are $8 \mathrm{cm}$ and $5 \mathrm{cm}$ respectively.

Chords of a circle $AB$ and $CD$ are $3 cm$and $4 cm$, which makes angle at centre are of $70°$ and $50°$ respectively.

In the figure below, $AB$ and $CD$are two equal chords of a circle and $O$ is the center of circle. If $OM\perp AB$ and $ON\perp CD$, then prove that $\angle OMN=\angle ONM$.

Prove that out of all chords which passes through any point of circle, that chord will be smallest which is perpendicular on diameter which passes through that point .