Class 9: Circles – Exercise 10.1

$\displaystyle \textbf{Question 1: }\text{The radius of a circle is }13\text{ cm and the length of one of its chords is}$
$\displaystyle 10\text{ cm. Find the distance of the chord from the centre.}$
$\displaystyle \text{Answer:}$ 2018-11-29_8-05-06 $\displaystyle \text{Let }AB\text{ be the chord and }OL\perp AB,\text{ where }O\text{ is the centre.}$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AL=LB=\frac{1}{2}AB=\frac{1}{2}\times10=5\text{ cm}$
$\displaystyle OA=13\text{ cm}$
$\displaystyle \text{In right-angled triangle }OLA,$
$\displaystyle OA^2=OL^2+AL^2$
$\displaystyle OL^2=OA^2-AL^2$
$\displaystyle OL=\sqrt{13^2-5^2}$
$\displaystyle =\sqrt{169-25}=\sqrt{144}=12\text{ cm}$
$\displaystyle \therefore \text{The distance of the chord from the centre is }12\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The radius of a circle is }8\text{ cm and the length of one of its chords is}$
$\displaystyle 12\text{ cm. Find the distance of the chord from the centre.}$
$\displaystyle \text{Answer:}$ 2018-11-29_8-05-17 $\displaystyle \text{Let }AB\text{ be the chord and }OL\perp AB,\text{ where }O\text{ is the centre.}$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AL=LB=\frac{1}{2}AB=\frac{1}{2}\times12=6\text{ cm}$
$\displaystyle OA=8\text{ cm}$
$\displaystyle \text{In right-angled triangle }OLA,$
$\displaystyle OA^2=OL^2+AL^2$
$\displaystyle OL^2=OA^2-AL^2$
$\displaystyle OL=\sqrt{8^2-6^2}$
$\displaystyle =\sqrt{64-36}=\sqrt{28}=2\sqrt{7}\text{ cm}$
$\displaystyle \therefore \text{The distance of the chord from the centre is }2\sqrt{7}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the length of a chord which is at a distance of }5\text{ cm from the}$
$\displaystyle \text{center of a circle of radius }13\text{ cm.}$
$\displaystyle \text{Answer:}$ 2018-11-29_19-58-20 $\displaystyle \text{Let }AB\text{ be the chord and }OL\perp AB,\text{ where }O\text{ is the centre.}$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AL=LB.$
$\displaystyle OA=13\text{ cm},\ OL=5\text{ cm}$
$\displaystyle \text{In right-angled triangle }OLA,$
$\displaystyle OA^2=OL^2+AL^2$
$\displaystyle AL^2=OA^2-OL^2$
$\displaystyle AL=\sqrt{13^2-5^2}$
$\displaystyle =\sqrt{169-25}=\sqrt{144}=12\text{ cm}$
$\displaystyle \therefore AB=2\times12=24\text{ cm}$
$\displaystyle \therefore \text{The length of the chord is }24\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the length of a chord which is at a distance of }5\text{ cm from the}$
$\displaystyle \text{center of a circle of radius }10\text{ cm.}$
$\displaystyle \text{Answer:}$ 2018-11-29_8-04-40 $\displaystyle \text{Let }AB\text{ be the chord and }OL\perp AB,\text{ where }O\text{ is the centre.}$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AL=LB.$
$\displaystyle OA=10\text{ cm},\ OL=5\text{ cm}$
$\displaystyle \text{In right-angled triangle }OLA,$
$\displaystyle OA^2=OL^2+AL^2$
$\displaystyle AL^2=OA^2-OL^2$
$\displaystyle AL=\sqrt{10^2-5^2}$
$\displaystyle =\sqrt{100-25}=\sqrt{75}=5\sqrt{3}\text{ cm}$
$\displaystyle \therefore AB=2\times5\sqrt{3}=10\sqrt{3}\text{ cm}$
$\displaystyle \therefore \text{The length of the chord is }10\sqrt{3}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the length of a chord which is at a distance of }4\text{ cm from the}$
$\displaystyle \text{center of a circle of radius }6\text{ cm.}$
$\displaystyle \text{Answer:}$ 2018-11-29_8-04-53 $\displaystyle \text{Let }AB\text{ be the chord and }OL\perp AB,\text{ where }O\text{ is the centre.}$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AL=LB.$
$\displaystyle OA=6\text{ cm},\ OL=4\text{ cm}$
$\displaystyle \text{In right-angled triangle }OLA,$
$\displaystyle OA^2=OL^2+AL^2$
$\displaystyle AL^2=OA^2-OL^2$
$\displaystyle AL=\sqrt{6^2-4^2}$
$\displaystyle =\sqrt{36-16}=\sqrt{20}=2\sqrt{5}\text{ cm}$
$\displaystyle \therefore AB=2\times2\sqrt{5}=4\sqrt{5}\text{ cm}$
$\displaystyle \therefore \text{The length of the chord is }4\sqrt{5}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Two chords }AB\text{ and }CD\text{ of lengths }5\text{ cm and }11\text{ cm,}$
$\displaystyle \text{respectively, of a circle are parallel. If the distance between }AB\text{ and }CD\text{ is}$
$\displaystyle 3\text{ cm, find the radius of the circle.}$
$\displaystyle \text{Answer:}$ 2018-11-29_8-04-07 $\displaystyle \text{Let }O\text{ be the centre and }OL\perp AB,\ OM\perp CD.$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AL=\frac{1}{2}AB=\frac{5}{2}=2.5\text{ cm}$
$\displaystyle \text{and }CM=\frac{1}{2}CD=\frac{11}{2}=5.5\text{ cm}.$
$\displaystyle \text{Let the radius of the circle be }r\text{ cm.}$
$\displaystyle \text{In right-angled triangle }OLA,$
$\displaystyle OL=\sqrt{OA^2-AL^2}=\sqrt{r^2-2.5^2}$
$\displaystyle \text{In right-angled triangle }OMC,$
$\displaystyle OM=\sqrt{OC^2-CM^2}=\sqrt{r^2-5.5^2}$
$\displaystyle \text{Since both chords lie on the same side of the centre, }OL-OM=3.$
$\displaystyle \sqrt{r^2-2.5^2}-\sqrt{r^2-5.5^2}=3$
$\displaystyle \sqrt{r^2-2.5^2}=3+\sqrt{r^2-5.5^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle r^2-2.5^2=9+r^2-5.5^2+6\sqrt{r^2-5.5^2}$
$\displaystyle 15=6\sqrt{r^2-5.5^2}$
$\displaystyle \sqrt{r^2-5.5^2}=\frac{5}{2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle r^2-5.5^2=\frac{25}{4}$
$\displaystyle r^2=\frac{25}{4}+\frac{121}{4}=\frac{146}{4}$
$\displaystyle r=\frac{\sqrt{146}}{2}\text{ cm}$
$\displaystyle \therefore \text{The radius of the circle is }\frac{\sqrt{146}}{2}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{An equilateral triangle of side }9\text{ cm is inscribed in a circle. Find the}$
$\displaystyle \text{radius of the circle.}$
$\displaystyle \text{Answer:}$ 2018-11-30_8-24-19 $\displaystyle \text{Let }O\text{ be the centre of the circle and }OL\perp BC.$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore BL=LC=\frac{1}{2}BC=\frac{9}{2}=4.5\text{ cm}$
$\displaystyle \text{Since }ABC\text{ is equilateral, }AL\text{ is also perpendicular to }BC.$
$\displaystyle \text{In right-angled triangle }ABL,$
$\displaystyle AL=\sqrt{AB^2-BL^2}$
$\displaystyle =\sqrt{9^2-4.5^2}=\sqrt{81-20.25}=\sqrt{60.75}$
$\displaystyle \text{Let the radius of the circle be }r\text{ cm.}$
$\displaystyle OA=OB=r\text{ and }OL=AL-OA=\sqrt{60.75}-r.$
$\displaystyle \text{In right-angled triangle }OBL,$
$\displaystyle OB^2=OL^2+BL^2$
$\displaystyle r^2=(\sqrt{60.75}-r)^2+4.5^2$
$\displaystyle r^2=60.75+r^2-2r\sqrt{60.75}+20.25$
$\displaystyle 2r\sqrt{60.75}=81$
$\displaystyle r=\frac{81}{2\sqrt{60.75}}$
$\displaystyle =\frac{81}{2\times\frac{9\sqrt{3}}{2}}=\frac{9}{\sqrt{3}}=3\sqrt{3}\text{ cm}$
$\displaystyle \therefore \text{The radius of the circle is }3\sqrt{3}\text{ cm}\approx5.196\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{AB is a diameter of a circle. }M\text{ is a point on }AB\text{ such that}$
$\displaystyle AM=18\text{ cm and }MB=8\text{ cm. Find the length of the shortest chord through }M.$
$\displaystyle \text{Answer:}$ 2018-11-29_8-03-25 $\displaystyle AB=AM+MB=18+8=26\text{ cm}$
$\displaystyle \therefore \text{Radius of the circle}=\frac{AB}{2}=\frac{26}{2}=13\text{ cm}$
$\displaystyle \therefore AO=OC=13\text{ cm}$
$\displaystyle OM=AM-AO=18-13=5\text{ cm}$
$\displaystyle \text{The shortest chord through }M\text{ is perpendicular to }OM.$
$\displaystyle \text{Let }CD\text{ be this chord, such that }OM\perp CD.$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore CM=MD.$
$\displaystyle \text{In right-angled triangle }OMC,$
$\displaystyle OC^2=OM^2+CM^2$
$\displaystyle CM=\sqrt{OC^2-OM^2}$
$\displaystyle =\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ cm}$
$\displaystyle \therefore CD=2CM=2\times12=24\text{ cm}$
$\displaystyle \therefore \text{The length of the shortest chord through }M\text{ is }24\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The length of the common chord of two intersecting circles is }30\text{ cm.}$
$\displaystyle \text{If the radii of the two circles are }25\text{ cm and }17\text{ cm, find the distance}$
$\displaystyle \text{between their centres.}$
$\displaystyle \text{Answer:}$ 2018-11-29_8-03-09 $\displaystyle \text{Let }AB\text{ be the common chord of the two circles with centres }O\text{ and }O'.$
$\displaystyle \text{Let }OO'\text{ intersect }AB\text{ at }Q.$
$\displaystyle \text{The line joining the centres of two intersecting circles bisects their common chord}$
$\displaystyle \text{perpendicularly.}$
$\displaystyle \therefore AQ=QB=\frac{1}{2}AB=\frac{1}{2}\times30=15\text{ cm}$
$\displaystyle \text{In right-angled triangle }OQA,$
$\displaystyle OA^2=OQ^2+AQ^2$
$\displaystyle OQ=\sqrt{OA^2-AQ^2}$
$\displaystyle =\sqrt{17^2-15^2}=\sqrt{289-225}=\sqrt{64}=8\text{ cm}$
$\displaystyle \text{In right-angled triangle }O'QA,$
$\displaystyle O'A^2=O'Q^2+AQ^2$
$\displaystyle O'Q=\sqrt{O'A^2-AQ^2}$
$\displaystyle =\sqrt{25^2-15^2}=\sqrt{625-225}=\sqrt{400}=20\text{ cm}$
$\displaystyle \text{Since the centres lie on opposite sides of the common chord,}$
$\displaystyle OO'=OQ+QO'=8+20=28\text{ cm}$
$\displaystyle \therefore \text{The distance between the centres is }28\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{A rectangle with a side of length }4\text{ cm is inscribed in a circle of}$
$\displaystyle \text{diameter }5\text{ cm. Find the area of the rectangle.}$
$\displaystyle \text{Answer:}$ 2018-11-29_8-00-38 $\displaystyle \text{Let }ABCD\text{ be the rectangle, where }AB=4\text{ cm.}$
$\displaystyle \text{Since the rectangle is inscribed in the circle, its diagonal is a diameter of the circle.}$
$\displaystyle \therefore AC=5\text{ cm}$
$\displaystyle \text{In right-angled triangle }ABC,$
$\displaystyle AC^2=AB^2+BC^2$
$\displaystyle 5^2=4^2+BC^2$
$\displaystyle BC^2=25-16=9$
$\displaystyle \therefore BC=3\text{ cm}$
$\displaystyle \text{Area of the rectangle}=AB\times BC$
$\displaystyle =4\times3=12\text{ cm}^2$
$\displaystyle \therefore \text{The area of the rectangle is }12\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{The center of a circle of radius }13\text{ units is the point }(3,6).$
$\displaystyle P(7,9)\text{ is a point inside the circle. }APB\text{ is a chord of the circle such that}$
$\displaystyle AP=PB.\text{ Calculate the length of }AB.$
$\displaystyle \text{Answer:}$ 2018-11-29_8-00-23 $\displaystyle \text{Let }O(3,6)\text{ be the centre of the circle.}$
$\displaystyle OP=\sqrt{(7-3)^2+(9-6)^2}$
$\displaystyle =\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\displaystyle \text{Since }AP=PB,\ P\text{ is the midpoint of chord }AB.$
$\displaystyle \text{The line joining the centre to the midpoint of a chord is perpendicular to the chord.}$
$\displaystyle \therefore OP\perp AB.$
$\displaystyle \text{In right-angled triangle }OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle AP=\sqrt{OA^2-OP^2}$
$\displaystyle =\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ units}$
$\displaystyle \therefore AB=2AP=2\times12=24\text{ units}$
$\displaystyle \therefore \text{The length of the chord }AB\text{ is }24\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{PQ and }RS\text{ are two parallel chords of a circle whose centre is }O$
$\displaystyle \text{and radius is }10\text{ cm. If }PQ=16\text{ cm and }RS=12\text{ cm, find the distance}$
$\displaystyle \text{between }PQ\text{ and }RS,\text{ if they lie:}$
$\displaystyle \text{(i) on the same side of the centre }O$
$\displaystyle \text{(ii) on opposite sides of the centre }O.$
$\displaystyle \text{Answer:}$ 2018-11-29_8-00-01 $\displaystyle \text{Let }OL\perp PQ\text{ and }OM\perp RS.$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore PL=LQ=\frac{1}{2}PQ=\frac{1}{2}\times16=8\text{ cm}$
$\displaystyle \text{and }RM=MS=\frac{1}{2}RS=\frac{1}{2}\times12=6\text{ cm.}$

$\displaystyle \text{(i) When the chords lie on the same side of the centre:}$
$\displaystyle \text{In right-angled triangle }OLP,$
$\displaystyle OP^2=OL^2+PL^2$
$\displaystyle OL=\sqrt{OP^2-PL^2}$
$\displaystyle =\sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\text{ cm}$
$\displaystyle \text{In right-angled triangle }OMR,$
$\displaystyle OR^2=OM^2+RM^2$
$\displaystyle OM=\sqrt{OR^2-RM^2}$
$\displaystyle =\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\text{ cm}$
$\displaystyle \therefore \text{Distance between the chords}=OM-OL=8-6=2\text{ cm}$
$\displaystyle \therefore \text{The distance between the chords is }2\text{ cm.}$

$\displaystyle \text{(ii) When the chords lie on opposite sides of the centre:}$
$\displaystyle \text{Distance between the chords}=OL+OM$
$\displaystyle =6+8=14\text{ cm}$
$\displaystyle \therefore \text{The distance between the chords is }14\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{AB and }CD\text{ are two parallel chords of a circle such that }AB=10\text{ cm}$
$\displaystyle \text{and }CD=24\text{ cm. If the chords are on opposite sides of the center and the distance}$
$\displaystyle \text{between them is }17\text{ cm, find the radius of the circle.}$
$\displaystyle \text{Answer:}$ 2018-11-29_7-59-31 $\displaystyle \text{Let }O\text{ be the center and }OL\perp AB,\ OM\perp CD.$
$\displaystyle \text{The perpendicular from the center of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AL=LB=\frac{1}{2}AB=\frac{1}{2}\times10=5\text{ cm}$
$\displaystyle \text{and }CM=MD=\frac{1}{2}CD=\frac{1}{2}\times24=12\text{ cm.}$
$\displaystyle \text{Let the radius of the circle be }r\text{ cm.}$
$\displaystyle \text{In right-angled triangle }OLA,$
$\displaystyle OL=\sqrt{OA^2-AL^2}=\sqrt{r^2-5^2}$
$\displaystyle \text{In right-angled triangle }OMC,$
$\displaystyle OM=\sqrt{OC^2-CM^2}=\sqrt{r^2-12^2}$
$\displaystyle \text{Since the chords lie on opposite sides of the center, }OL+OM=17.$
$\displaystyle \sqrt{r^2-5^2}+\sqrt{r^2-12^2}=17$
$\displaystyle \sqrt{r^2-5^2}=17-\sqrt{r^2-12^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle r^2-25=289+r^2-144-34\sqrt{r^2-144}$
$\displaystyle 34\sqrt{r^2-144}=170$
$\displaystyle \sqrt{r^2-144}=5$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle r^2-144=25$
$\displaystyle r^2=169$
$\displaystyle r=13\text{ cm}$
$\displaystyle \therefore \text{The radius of the circle is }13\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{AB and }CD\text{ are two chords of a circle such that }AB=6\text{ cm}$
$\displaystyle \text{and }CD=12\text{ cm. }AB\parallel CD.\text{ If the distance between }AB\text{ and }CD\text{ is}$
$\displaystyle 3\text{ cm, find the radius of the circle.}$
$\displaystyle \text{Answer:}$ 2018-11-29_7-59-44 $\displaystyle \text{Let }O\text{ be the center and }OM\perp AB,\ OL\perp CD.$
$\displaystyle \text{The perpendicular from the center of a circle to a chord bisects the chord.}$
$\displaystyle \therefore AM=MB=\frac{1}{2}AB=\frac{1}{2}\times6=3\text{ cm}$
$\displaystyle \text{and }CL=LD=\frac{1}{2}CD=\frac{1}{2}\times12=6\text{ cm.}$
$\displaystyle \text{Let the radius of the circle be }r\text{ cm.}$
$\displaystyle \text{In right-angled triangle }OMA,$
$\displaystyle OM=\sqrt{OA^2-AM^2}=\sqrt{r^2-3^2}$
$\displaystyle \text{In right-angled triangle }OLC,$
$\displaystyle OL=\sqrt{OC^2-CL^2}=\sqrt{r^2-6^2}$
$\displaystyle \text{Since both chords lie on the same side of the center, }OM-OL=3.$
$\displaystyle \sqrt{r^2-3^2}-\sqrt{r^2-6^2}=3$
$\displaystyle \sqrt{r^2-3^2}=3+\sqrt{r^2-6^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle r^2-9=9+r^2-36+6\sqrt{r^2-36}$
$\displaystyle 18=6\sqrt{r^2-36}$
$\displaystyle \sqrt{r^2-36}=3$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle r^2-36=9$
$\displaystyle r^2=45$
$\displaystyle r=\sqrt{45}=3\sqrt{5}\text{ cm}$
$\displaystyle \therefore \text{The radius of the circle is }3\sqrt{5}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{ABC is an isosceles triangle inscribed in a circle. If }AB=AC=15\text{ cm}$
$\displaystyle \text{and }BC=18\text{ cm, find the radius of the circle.}$
$\displaystyle \text{Answer:}$ 2018-11-29_7-58-51 $\displaystyle \text{Let }O\text{ be the centre of the circle and }OM\perp BC.$
$\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}$
$\displaystyle \therefore BM=MC=\frac{1}{2}BC=\frac{1}{2}\times18=9\text{ cm}$
$\displaystyle \text{Since }AB=AC,\ AM\text{ is also perpendicular to }BC.$
$\displaystyle \text{In right-angled triangle }ABM,$
$\displaystyle AM=\sqrt{AB^2-BM^2}$
$\displaystyle =\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12\text{ cm}$
$\displaystyle \text{Let the radius of the circle be }r\text{ cm.}$
$\displaystyle OA=OB=r.$
$\displaystyle \text{In right-angled triangle }OMB,$
$\displaystyle OM=\sqrt{OB^2-BM^2}=\sqrt{r^2-9^2}$
$\displaystyle \text{Since }A,O\text{ and }M\text{ are collinear and }O\text{ lies between }A\text{ and }M,$
$\displaystyle AM=AO+OM$
$\displaystyle 12=r+\sqrt{r^2-9^2}$
$\displaystyle \sqrt{r^2-9^2}=12-r$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle r^2-81=144+r^2-24r$
$\displaystyle 24r=225$
$\displaystyle r=\frac{225}{24}=\frac{75}{8}=9\frac{3}{8}\text{ cm}$
$\displaystyle \therefore \text{The radius of the circle is }\frac{75}{8}\text{ cm or }9\frac{3}{8}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{In a circle of radius }5\text{ cm, }AB\text{ and }AC\text{ are two chords such that}$
$\displaystyle AB=AC=6\text{ cm. Find the length of chord }BC.$
$\displaystyle \text{Answer:}$ 2018-11-29_7-59-03 $\displaystyle \text{Let }O\text{ be the centre of the circle and }OM\perp BC.$
$\displaystyle \text{Since }AB=AC,\text{ equal chords are equidistant from the centre.}$
$\displaystyle \therefore AO\text{ is the perpendicular bisector of }BC.$
$\displaystyle \therefore BM=MC.$
$\displaystyle \text{Let }BM=MC=x\text{ cm.}$
$\displaystyle \text{In right-angled triangle }OMB,$
$\displaystyle OM=\sqrt{OB^2-BM^2}=\sqrt{5^2-x^2}=\sqrt{25-x^2}$
$\displaystyle \text{Since }M\text{ lies between }A\text{ and }O,$
$\displaystyle AM=AO-OM=5-\sqrt{25-x^2}$
$\displaystyle \text{In right-angled triangle }AMB,$
$\displaystyle AB^2=AM^2+BM^2$
$\displaystyle 6^2=\left(5-\sqrt{25-x^2}\right)^2+x^2$
$\displaystyle 36=25+25-x^2-10\sqrt{25-x^2}+x^2$
$\displaystyle 36=50-10\sqrt{25-x^2}$
$\displaystyle 10\sqrt{25-x^2}=14$
$\displaystyle \sqrt{25-x^2}=\frac{7}{5}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle 25-x^2=\frac{49}{25}$
$\displaystyle x^2=25-\frac{49}{25}=\frac{576}{25}$
$\displaystyle x=\frac{24}{5}=4.8\text{ cm}$
$\displaystyle \therefore BC=BM+MC=2x$
$\displaystyle =2\times4.8=9.6\text{ cm}$
$\displaystyle \therefore \text{The length of chord }BC\text{ is }9.6\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Two concentric circles with center }O\text{ have }A,B,C,D\text{ as the points of}$
$\displaystyle \text{intersection with the line }l\text{ as shown in the diagram. If }AD=12\text{ cm and }BC=8\text{ cm,}$
$\displaystyle \text{find the lengths of }AB,\ CD,\ AC\text{ and }BD.$
$\displaystyle \text{Answer:}$ 2018-11-29_7-58-29 $\displaystyle \text{Since the line }l\text{ passes through the common center }O,\ AD\text{ and }BC\text{ are diameters.}$
$\displaystyle OA=OD=\frac{1}{2}AD=\frac{1}{2}\times12=6\text{ cm}$
$\displaystyle OB=OC=\frac{1}{2}BC=\frac{1}{2}\times8=4\text{ cm}$
$\displaystyle AB=OA-OB=6-4=2\text{ cm}$
$\displaystyle CD=OD-OC=6-4=2\text{ cm}$
$\displaystyle AC=AB+BC=2+8=10\text{ cm}$
$\displaystyle BD=BC+CD=8+2=10\text{ cm}$
$\displaystyle \therefore AB=2\text{ cm},\ CD=2\text{ cm},\ AC=10\text{ cm and }BD=10\text{ cm}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Two circles of radii }10\text{ cm and }8\text{ cm intersect, and the length of}$
$\displaystyle \text{their common chord is }12\text{ cm. Find the distance between their centers.}$
$\displaystyle \text{Answer:}$ 2018-11-29_7-57-49 $\displaystyle \text{Let }AB\text{ be the common chord of the circles with centers }O\text{ and }O'.$
$\displaystyle \text{Let }OO'\text{ intersect }AB\text{ at }Q.$
$\displaystyle \text{The line joining the centers of two intersecting circles bisects their common chord}$
$\displaystyle \text{perpendicularly.}$
$\displaystyle \therefore AQ=QB=\frac{1}{2}AB=\frac{1}{2}\times12=6\text{ cm}$
$\displaystyle \text{In right-angled triangle }OQA,$
$\displaystyle OQ=\sqrt{OA^2-AQ^2}$
$\displaystyle =\sqrt{8^2-6^2}=\sqrt{64-36}=\sqrt{28}=2\sqrt{7}\text{ cm}$
$\displaystyle \text{In right-angled triangle }O'QA,$
$\displaystyle O'Q=\sqrt{O'A^2-AQ^2}$
$\displaystyle =\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\text{ cm}$
$\displaystyle \text{Since the centers lie on opposite sides of the common chord,}$
$\displaystyle OO'=OQ+QO'=2\sqrt{7}+8$
$\displaystyle =8+2\sqrt{7}\text{ cm}\approx13.29\text{ cm}$
$\displaystyle \therefore \text{The distance between the centers is }8+2\sqrt{7}\text{ cm, or approximately }13.29\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{In the figure, two circles with centers }A\text{ and }B\text{ and radii }5\text{ cm}$
$\displaystyle \text{and }3\text{ cm touch each other internally. If the perpendicular bisector of segment }AB$
$\displaystyle \text{meets the bigger circle at }P\text{ and }Q,\text{ find the length of }PQ.$
$\displaystyle \text{Answer:}$ 2018-11-29_7-57-28 $\displaystyle \text{Let the perpendicular bisector of }AB\text{ intersect }AB\text{ at }M.$
$\displaystyle \text{Since the two circles touch internally,}$
$\displaystyle AB=5-3=2\text{ cm}$
$\displaystyle \text{Since }M\text{ is the midpoint of }AB,$
$\displaystyle AM=MB=\frac{1}{2}AB=\frac{1}{2}\times2=1\text{ cm}$
$\displaystyle \text{Also, }PQ\perp AB\text{ at }M.$
$\displaystyle \text{The perpendicular from the center of a circle to a chord bisects the chord.}$
$\displaystyle \therefore PM=MQ.$
$\displaystyle \text{In right-angled triangle }AMP,$
$\displaystyle AP^2=AM^2+PM^2$
$\displaystyle PM=\sqrt{AP^2-AM^2}$
$\displaystyle =\sqrt{5^2-1^2}=\sqrt{25-1}=\sqrt{24}=2\sqrt{6}\text{ cm}$
$\displaystyle \therefore PQ=2PM=2\times2\sqrt{6}=4\sqrt{6}\text{ cm}$
$\displaystyle \therefore \text{The length of }PQ\text{ is }4\sqrt{6}\text{ cm.}$
$\displaystyle \\$