Note: In almost all the problems below the implementation of the following theorem is important. Theorem 3: The perpendicular from the center of a circle to a chord bisects the chord.

Question 1: The radius of the circle is $13 \ cm$ and the length of one of its chords is $10 \ cm$. Find the distance of the chord from the center.

$OL = \sqrt{OA^2 - AL^2} = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm

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Question 2: The radius of the circle is $8 \ cm$ and the length of one of its chords is $12 \ cm$. Find the distance of the chord from the center.

$OL = \sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 2\sqrt{7}$ cm

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Question 3: Find the length of a chord which is at a distance of $5 \ cm$ from the center of the circle of radius $13 \ cm$.

$AL = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm

Therefore $AB = 2 \times 12 = 24$ cm

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Question 4: Find the length of a chord which is at a distance of $5 \ cm$ from the center of the circle of radius $10 \ cm$.

$AL = \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3}$ cm

Therefore $AB = 2 \times 5\sqrt{3} = 10\sqrt{3}$ cm

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Question 5: Find the length of a chord which is at a distance of $4 \ cm$ from the center of the circle of radius $6 \ cm$.

$AL = \sqrt{6^2 - 4^2} = \sqrt{36-16} = \sqrt{20} = 2\sqrt{5}$ cm

Therefore $AB = 2 \times 2\sqrt{5} = 4\sqrt{5}$ cm

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Question 6: Two chords $AB$ and $CD$ of length $5 \ cm$ and $11 \ cm$ respectively of a circle are parallel. If the distance between $AB$ and $CD$ is $3 \ cm$, find the radius of the circle.

$OM = \sqrt{r^2 - 5.5^2}$

$OL = \sqrt{r^2 - 2.5^2}$

$OL - OM = 3$

$\sqrt{r^2 - 2.5^2} - \sqrt{r^2 - 5.5^2} = 3$

$\Rightarrow \sqrt{r^2 - 2.5^2} = 3 + \sqrt{r^2 - 5.5^2}$

Squaring both sides we get

$r^2 - 2.5^2 = 9 + r^2 - 5.5^2 + 6 \sqrt{r^2 - 5.5^2}$

$\Rightarrow 15 = 6 \sqrt{r^2 - 5.5^2}$

$\Rightarrow 5 = 2 \sqrt{r^2 - 5.5^2}$

Squaring both sides we get

$25 = 4(r^2 - 5.5^2)$

$\Rightarrow r^2 =$ $\frac{25}{4}$ $+ 5.5^2$

$\Rightarrow r = \sqrt{\frac{25}{4} + 5.5^2} =$ $\sqrt{\frac{146}{4}}$ $=$ $\frac{\sqrt{146}}{2}$

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Question 7: An equilateral triangle of side $9 \ cm$ is inscribed in a circle. Find the radius of the circle.

$AL = \sqrt{9^2 - 4.5^2} = \sqrt{81-20.05} = \sqrt{60.75}$

$OL = \sqrt{60.75} - r$

$OB^2 = OL^2 + BL^2$

$r^2 = (\sqrt{60.75} - r)^2 + 4.5^2$

$\Rightarrow r^2 = 60.75 + r^2 - 2\sqrt{60.75} - r + 4.5^2$

$\Rightarrow 2 \sqrt{60.75} \ r = 60.75 + 4.5^2$

$\Rightarrow 2 \sqrt{60.75} \ r = 81$

$\Rightarrow r =$ $\frac{81}{2\sqrt{60.75}}$ $= 5.196$ cm

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Question 8: $AB$ is a diameter of the circle. $M$ is a point in $AB$ such that $AM = 18 \ cm$ and $MB = 8 \ cm$. Find the length of the shortest chord through $M$.

$r =$ $\frac{18+8}{2}$ $= 13$ cm

$AM = 18$ cm

$AO = 13$ cm

Therefore $OM = 5$

Therefore $MC = \sqrt{13^2- 5^2} = 12$ cm

Hence $CD = 2 \times 12 = 24$ cm

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Question 9: The length of the common chord of two intersecting circles is $30 \ cm$. If the radii of the two circles are $25 \ cm$ and $17 \ cm$, find the distance between their centers.

$QR = r_1 \ \ \ \ \ OA = 17 \ \ \ \ \ OQ = 17-r_1 \ \ \ \ \ AQ = 15$

Therefore $17^2 = (17-r_1)^2 + 15^2$

$289 - 225 = (17-r_1)^2$

$\Rightarrow 64 = (17-r_1)^2$

$\Rightarrow 8 = 17 - r_1$

$\Rightarrow r_1 = 9$ cm

Similarly, $PQ = r_2$

$O'B = 25 \ \ \ \ \ QB = 15$

Therefore $QO' = 25 - r_2$

$\Rightarrow 25^2 = (25-r_2)^2 + 15^2$

$\Rightarrow 625-225 = (25-r_2)^2$

$\Rightarrow 20 = 25 - r_2$

$\Rightarrow r_2 = 5$ cm

Hence $OO' = OR + PO' - r_1 -r_2$

$\Rightarrow OO'= 17 + 25 - 9 - 5$

$\Rightarrow OO'= 28$ cm

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Question 10: A rectangle with a side of length $4 \ cm$ is inscribed in a circle of diameter $5 \ cm$. Find the area of the rectangle.

$OL^2 = 2.5^2 - 2^2 = 6.25 - 4 = 2.25$

$\Rightarrow OL = 1.5$ cm

$AD = 2 \times 1.5 = 3$ cm

Hence the area of rectangle $= 4 \times 3 = 12 \ cm^2$

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Question 11: The center of a circle of radius $13 \ cm$ units is the point $(3,6)$. $P(7,9)$ is a point inside the circle. $APB$ is a chord of the circle such that $AP = PB$. Calculate the length of $AB$.

$OP = \sqrt{(9-6)^2+(7-3)^2} = \sqrt{9+16} = \sqrt{25} = 5$ cm

Therefore $AP = \sqrt{13^2 - 5^2} = 12$ cm

Hence $AB = 2 \times 12 = 24$ cm

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Question 12: $PQ$ and $RS$ are two parallel chords of a circle whose center is $O$ and radius is $10 \ cm$. If $PQ$ is $16 \ cm$ and $RS$ is $12 \ cm$, find the distance between $PQ$ and $RS$ , if they lie i) on the same side of center $O$ ii) on the opposite side of center $O$

i) $OL = \sqrt{10^2 - 8^2} = \sqrt{36} = 6$ cm

$OM = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ cm

Therefore $LM = 8 - 6 = 2$ cm

ii) $OL = \sqrt{10^2 - 8^2} = \sqrt{36} = 6$ cm

$OM = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ cm

Therefore $LM = 8 + 6 = 14$ cm

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Question 13: $AB$ and $CD$ are two parallel chords of a circle such that $AB = 10 \ cm$ and $CD = 24 \ cm$. If the chords are on the opposite sides of the center and the distance between then is $17 \ cm$, find the radius of the circle.

$OL^2 = r^2 - 5^2$

$\Rightarrow OL = \sqrt{r^2 -5^2}$

Similarly, $OM = \sqrt{r^2 - 12^2}$

$OL + OM = 17$

Therefore $\sqrt{r^2 -5^2} + \sqrt{r^2 - 12^2} = 17$

$\Rightarrow \sqrt{r^2 -5^2} = 17 - \sqrt{r^2 - 12^2}$

Squaring both sides

$r^2 - 5^2 = 289 + r^2 - 12^2 -34\sqrt{r^2 - 12^2}$

$\Rightarrow 34\sqrt{r^2 - 12^2} = 289 - 144 + 25$

$\Rightarrow 34\sqrt{r^2 - 12^2} = 170$

$\Rightarrow \sqrt{r^2 - 12^2} = 5$

Squaring both sides

$r^2 - 12^2 = 25$

$\Rightarrow r^2 = 144 + 25$

$\Rightarrow r = 13$ cm

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Question 14: $AB$ and $CD$ are two chords of a circle such that $AB = 6 \ cm$ and $CD = 12 \ cm$. $AB \parallel CD$. If the distance between $AB$ and $CD$ is $3 \ cm$, find the radius of the circle.

$OM = \sqrt{r^2 - 6^2}$

$OL = \sqrt{r^2 - 3^2}$

$OL - OM = 3$

$\sqrt{r^2 - 3^2} - \sqrt{r^2 - 6^2} = 3$

$\Rightarrow \sqrt{r^2 - 3^2} = 3 + \sqrt{r^2 - 6^2}$

Squaring both sides

$r^2 - 3^2 = 9 + r^2 - 6^2 + 6\sqrt{r^2 - 6^2}$

$\Rightarrow 18 = 6 \sqrt{r^2 - 6^2}$

$\Rightarrow 3 = \sqrt{r^2 - 6^2}$

Squaring both sides

$9 = r^2 - 6^2$

$\Rightarrow r^2 = 45$

$\Rightarrow r = \sqrt{45}$ cm

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Question 15: $ABC$ is an isosceles triangle inscribed in a circle. If $AB = AC = 15 \ cm$ and $BC = 18 \ cm$, find the radius of the circle.

$OM = \sqrt{r^2 - 9^2}$

$AM = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = 12$ cm

$r + \sqrt{r^2 - 9^2} = 12$

$\Rightarrow \sqrt{r^2 - 9^2} = 12 - r$

Squaring both sides

$r^2 - 9^2 = 144 + r^2 - 24 r$

$\Rightarrow 24r = 225$

$\Rightarrow r =$ $\frac{75}{8}$ $= 9$ $\frac{3}{8}$ cm

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Question 16: In a circle of radius $5 \ cm$, $AB$ and $AC$ are two chords such that $AB = AC = 6 \ cm$. Find the length of the chord $BC$.

$OM = \sqrt{5^2 - x^2}$

$AM = 5 + OM = 5 + \sqrt{5^2 - x^2}$

Therefore $(5 + \sqrt{5^2 - x^2})^2 + x^2 = 6^2$

$\Rightarrow 25 + 5^2 - x^2 + 10 \sqrt{5^2 - x^2} = 6^2$

$\Rightarrow 50 + 10 \sqrt{5^2 - x^2} = 36$

$\Rightarrow 10 \sqrt{5^2 - x^2} = -14$

Squaring both sides

$100 (5^2 - x^2) = 196$

$\Rightarrow x^2 = 25 - 1.96 = \sqrt{23.04} = 4.79$ cm

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Question 17: Two concentric circles with center $O$ have $A, B, C, D$ as the points of intersection with the line $l$ as shown in the diagram. If $AD = 12 \ cm$ and $BC = 8 \ cm$, find the lengths of $AB, CD, AC$ and $BD$

$AD = 12$ cm    $BC = 8$ cm

$AB = 2$ cm      $CD = 2$ cm

$AC = 2 + 8 = 10$ cm

$BD = 8 + 2 = 10$ cm

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Question 18: Two circles of radii $10 \ cm$ and $8 \ cm$ intersect and the length of the common chord is $12 \ cm$. Find the distance between their centers.

$(8-r_1)^2 + 6^2 = 8^2$

$\Rightarrow (8-r_1)^2 = 64 - 36 = 28$

$\Rightarrow 8 - r_1 = 2\sqrt{7}$

$\Rightarrow r_1 = 8 - 2\sqrt{7} = 2.71$

Similarly, $(10 - r_2)^2 + 6^2 = 10^2$

$\Rightarrow (10 - r_2)^2 = 100 - 36 = 64$

$\Rightarrow 10 - r_2 = 8$

$\Rightarrow r_2 = 2$

Therefore $OO' = (8-r_1) + (10 - r_2)$

$\Rightarrow OO'= 8 - 2.71 + 10 -2$

$\Rightarrow OO'= 16 - 2.71 = 13.29$ cm

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Question 19: In the figure, two circles with center $A$ and $B$ and of radii $5 \ cm$ and $3 \ cm$ touch each other internally. If the perpendicular bisector of segment $AB$ meets the bigger circle in $P$ and $Q$ , find the length of $PQ$.

$PB = \sqrt{5^2 - 1^2} = \sqrt{24} = 2\sqrt{6}$
Therefore $PQ = 2 \times 2\sqrt{6} = 4\sqrt{6}$ cm
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