NCERT Solutions for Class 10 Math Chapter 10 - Circles

Answer:

Since at any point on a circle, there can be one and only one tangent. But circle is a collection of infinite points, so we can draw infinite number of tangents to a circle.

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one

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secant

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two

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point of contact

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(iv)

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According to the given information we draw a circle having O as centre and l is the given line.

Now, m and n be two lines parallel to a given line l such that m is tangent as well as parallel to l and n is secant to the circle as well as parallel to l.

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(i) 7 cm

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(ii) 70$°$

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(i) 50$°$

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Given: A circle with centre O. Its tangent AB meets circle at P. i.e., P is the point of contact. To prove: Perpendicular at the point of contact to the tangent to a circle passes through the centre.

Construction: Join OP. Proof: The perpendicular to a tangent line AB through the point of contact passes through the centre of the circle because only one perpendicular, OP can be drawn to the line AB through the point P.

Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre.

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A circle with centre ‘O’. A is any point outside the circle at a distance of 5 cm from the centre.

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Two concentric circles having same centre O, and radii 5 cm and 3 cm respectively. Let PQ be the chord of larger circle but tangent to the smaller circle.

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Given: A quadrilateral ABCD is drawn to circumscribe a circle having O as its centre. To prove: AB + CD = AD + BC Proof: Since, the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside the circle and BQ; BP are tangents to the circle.
∴ BQ = BP ...(1)
Similarly, AS = AP ...(2)
and CQ = CR ...(3)
Also, DS = DR ...(4)
Adding (1), (2), (3) and (4), we get
(BQ + CQ) + (AS + DS) = (BP + AP) + (CR + DR)
BC + AD = AB + CD
is the required result.

Answer:

Given: A circle with centre O having two parallel tangents XY and X′Y′, another tangent AB with point of contact C intersecting XY at A and X′Y′ at B.
To prove: $\angle$AOB = 90$°$
Construction: Join OC, OA and OB.
Proof: Since, the lengths of tangents drawn from an external point to a circle are equal.
Now, A is any point outside the circle from two tangents PA and CA are drawn to the circle.
∴ PA = CA
Also, in $∆$ POA and $∆$ AOC,
PA = CA (Proved)
OA = OA (Common side)
OP = OC (Radii of same circle)
∴ $∆$ POA $\cong$ $∆$ AOC [SSS congruence]
and $\angle$PAO = $\angle$CAO [CPCT]
or $\angle$PAC = 2 $\angle$PAO = 2 $\angle$CAO ...(1)
Similarly,
$\angle$QBC = 2$\angle$OBC = 2 $\angle$OBQ ...(2)
Now, $\angle$PAC + $\angle$QBC = 180$°$
[Sum of the interior angles on the same side of
transversal is 180$°$.]

Answer:

Given: A circle with centre O. P is any point outside the circle from PQ and PR are the tangents to the given circle.

To prove: $\angle$ROQ + $\angle$QPR = 180$°$
Proof: OQ is the radius and PQ is tangent from point P to the given circle.
∴ $\angle$OQP = 90$°$ ...(1)
[The tangent at any point of a circle is perpendicular to the radius through the point of contact.]
Similarly, $\angle$ORP = 90$°$ ...(2) Now, in quadrilateral ROQP,
$\angle$ROQ + $\angle$PRO + $\angle$OQP + $\angle$QPR = 360$°$ or $\angle$ROQ + 90$°$ + 90$°$ + $\angle$QPR = 360$°$ [Using (1) & (2)]
or $\angle$ROQ + $\angle$QPR + 180 = 360$°$ or $\angle$ROQ + $\angle$QPR = 360$°$ – 180$°$ = 180$°$ Hence, the angle between the two tangents drawn from an external point to a circle is supplementary to angle subtended by the line segment joining the points of contact at the centre.

Answer:

Given: A parallelogram ABCD circumscribing a circle with centre O.

To prove: ABCD is a rhombus.
Proof: We know that, the lengths of tangents
drawn from an external point to a circle are equal.
Now, B is any point outside the circle and BE,
BF are tangents to the circle.
∴ BE = BF ...(1)
Similarly, AE = AH ...(2)
and CG = CF ...(3)
Also DG = DH ...(4)
Adding (1), (2), (3) and (4), we get
(BE + AE) + (CG + DG) = (BF + CF) + (AH + DH)
or AB + CD = BC + AD ...(5)
Now, ABCD is a parallelogram,
∴ AB = CD and BC = AD ...(6)
From (5) and (6), we get
AB + AB = BC + BC
or 2AB = 2BC or AB = BC
or AB = BC = CD = AD
∴ ABCD is a rhombus.
Hence, parallelogram circumscribing a circle is a rhombus.

Answer:

A triangle ABC is drawn to circumscribe a circle of radius 4 cm and the sides BC, CA, AB of ΔABC touch the circle at D, E, F respectively. Since, the lengths of tangents drawn from an external point to a circle are equal.
∴ AE = AF = x cm (say)
CE = CD = 6 cm
and BF = BD = 8 cm
Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Answer:

Given: A quadrilateral PQRS circumscribing a circle having centre O. Sides PQ, QR, RS and SP touch the circle at L, M, N, T respectively.

To prove: $\angle$POQ + $\angle$SOR = 180$°$
and $\angle$SOP + $\angle$ROQ = 180$°$
Construction: Join OP, OL, OQ, OM, OR, ON, OS, OT
Proof: Since, the two tangents drawn from an external point subtend equal angles at the centre.
∴ $\angle$2 = $\angle$3; $\angle$4 = $\angle$5 ; $\angle$6 = $\angle$7; $\angle$8 = $\angle$1 ...(1)
But, sum of all angles around a point is 360$°$
∴ $\angle$1 + $\angle$2 + $\angle$3 + $\angle$4 + $\angle$5
+ $\angle$6 + $\angle$7 + $\angle$8 = 360$°$
or $\angle$1 + $\angle$2 + $\angle$2 + $\angle$5 + $\angle$5
+ $\angle$6 + $\angle$6 + $\angle$1 = 360$°$