Free Circles 02 Practice Test - 10th Grade
Question 1
The length of a tangent from a point Q to a circle is 24 cm. The distance between Q and the center of the circle is 25 cm. The radius of the circle is
SOLUTION
Solution :
Given that PQ = 24 cm and OQ = 25 cm.
To find OP, apply Pythagoras theorem for △OPQ.
OP2+QP2=OQ2
OP2+242=252
OP2=625−576=49
⇒OP=7 cm
Question 2
State true or false.
PQ is a tangent drawn from a point P to a circle with center O and QOR is a diameter of the circle such that ∠POR = 120∘ , then ∠OPQ is 30∘.
True
False
SOLUTION
Solution : A
Given that ∠POR=120∘
We know that ∠OQP is 90∘
Through external angle theorem ∠POR=∠OQP+∠OPQ
⇒120∘=90∘+∠OPQ
∠OPQ=120∘–90∘
∠OPQ=30∘
Question 3
Three circles touch each other externally. The distance between their centres is 5 cm, 6 cm and 7 cm. Find the radii of the circles.
2 cm, 3 cm, 4 cm
3 cm, 4 cm, 1 cm
1 cm, 2.5 cm, 3.5 cm
1 cm, 2 cm, 4 cm
SOLUTION
Solution : A
Consider the below figure wherein three circles touch each other externally.
Since the distances between the centres of these circles are 5 cm, 6 cm and 7 cm respectively, we have the following set of equations with respect to the above diagram:
x+y = 5 …..(1)y+z = 6 ......(2) (⇒ y=6-z)... (2.1)
x+z = 7 …..(3)Adding (1), (2) and (3), we have 2(x+y+z)=5+6+7=18
⟹x+y+z=9....(4)
Using (1) in (4), we have 5+z=9⟹z=4
Now using, (3)⟹x=7−z=7−4=3
And (2.1)⟹y=6−z=6−4=2Therefore, the radii of the circles are 3 cm, 2 cm and 4 cm.
Question 4
In the given figure, a circle touches the side BC of ΔABC at P and AB and AC produced at Q and R respectively. If AQ = 15 cm, find the perimeter of ΔABC.
SOLUTION
Solution :We know that
AQ = AR
AQ = AC + CR
But CR = CP
Therefore AQ = AC + CP …..(i)
Also AQ = AB + BQ
But BQ = BP
∴ AQ = AB + BP .….(ii)
Adding equations (i) and (ii)
2AQ = AC + CP + BP + AB
2AQ = Perimeter of ΔABC
Perimeter of ΔABC = 2 × 15 = 30 cm
Question 5
Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.
√a2−b2
√a2+b2
2√a2−b2
2√a2+b2
SOLUTION
Solution : C
Let O be the common centre of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.
Join OA and OC.
Then OC ⊥ AB
Let OA = a and OC = b.
Since OC ⊥ AB, OC bisects AB
[ ∵ perpendicular from the centre to a chord bisects the chord].In right Δ ACO, we have
OA2=OC2+AC2 [by Pythagoras' theorem]
⇒ AC=√OA2−OC2=√a2−b2
∴ AB=2AC=2√a2−b2 [ ∵ C is the midpoint of AB]i.e., Length of the chord AB=2√a2−b2
Question 6
In the given figure, the circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R, S respectively. If AB = 11 cm, BC =x cm, CR = 4 cm and AS = 6 cm, the value of x is
SOLUTION
Solution :AP = AS = 6 cm
BP = BA – AP = 11 – 6 = 5
BQ = BP = 5
CQ = CR = 4
x = CQ + BQ
x = 4 + 5 = 9 cm
Question 7
ABC is an isosceles triangle and AC, BC are the tangents at M and N respectively. DE is the diameter of the circle. ∠ADP = ∠BEQ = 100∘. What is value of ∠PRD?
60∘
50∘
20∘
Can't be determined
SOLUTION
Solution : C
ADB is a straight line. By linear pair axiom,
∠ADP + ∠PDB = 180∘
100 + ∠PDB = 180∘
∠PDB = 80∘
Similarly ∠QED = 80∘
We have, ∠DPE = 90∘ (angle subtended by a diameter)In △DPE,
∠DPE + ∠PED + ∠EDP = 180
[Angle sum property of a triangle]
∠PED = 10∘
Similarly ∠QDE = 10∘In △DRE,
∠DRE + ∠RDE + ∠RED = 180
[Angle sum property of a triangle]
∠DRE = 160∘PRE is a straight line. By linear pair axiom,
∠PRD + ∠DRE = 180∘
∠PRD + 160 = 180∘
∠PRD = 20∘
Question 8
In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then
AD=AB+BC+CA
2AD=AB+BC+CA
3AD=AB+BC+CA
4AD=AB+BC+CA
SOLUTION
Solution : B
We know thatAD=AE
AD=AB+BE
Since BE=BF as tangents drawn from an external point to a circle are equal , AD=AB+BF……(1)
Also
AD=AC+CD
AD=AC+CF ……(2) (CD and CF are tangents drawn from the external point C to the circle)
Adding equation (1) and (2),
AD+AD=AB+BF+CF+AC
2AD=AB+BC+AC
Question 9
State True or False.
PA and PB are tangents to the circle with centre O from an external point P, touching the circle at A and B respectively. Then the quadrilateral AOBP is cyclic.
True
False
SOLUTION
Solution : A
Given PA and PB are tangents to the circle with centre O from an external point P.
We know that the tangent at any point of a circle is perpendicular to radius at the point of contact.
∴ PA⊥OA, i.e., ∠OAP=90∘ . . . (i)
and PB⊥OB, i.e., ∠OBP=90∘ . .. (ii)Now, the sum of all the angles of a quadrilateral is 360∘
∴ ∠AOB+∠OAP+∠APB+∠OBP=360∘
⇒ ∠AOB+∠APB=180∘ [using (i) and (ii)]∴ quadrilateral OAPB is cyclic [since both pairs of opposite angles have the sum 180∘].
Question 10
O is the centre of the circle and line XY shares one common point with the circle. A line segment OP is drawn from O to line XY such that P is a point on line XY. What happens if P lies on the circle?
This situation holds no signifiance
OP needs be the radius of circle
OP is perpendicular to XY
The angle between OP and XY is variable
SOLUTION
Solution : B and C
XY touches the circle at only one point. Thus, it is a tangent to the circle. The tangent touches the circle at only one point which lies on the circumference (point P).
Thus, any other point on line XY other than P lies outside the circle. Let such a point be Q. Now let us compare lengths OP and OQ. It is clear that OP is equal to radius since P lies on circle. Q lies outside. Thus, OP is lesser than OQ. A point which is outside the circle will obviously be farther away from circle centre than a point which lies on the circle.
It can be seen that at any position of point Q on line XY except on P, OP is lesser than OQ. In other words, OP is the smallest distance between O and XY. The smallest distance between a line and a point is the perpendicular distance. Hence OP is perpendicular to XY.