Free Circles 02 Practice Test - 10th Grade 

Given that PQ = 24 cm and OQ = 25 cm.

To find OP, apply Pythagoras theorem for $△$OPQ.

${OP}^2+{QP}^2={OQ}^2$

${OP}^2+{24}^2={25}^2$

${OP}^2=625-576=49$

$\Rightarrow OP=7\text{cm}$

Given that $\angle POR={120}^{\circ }$

We know that $\angle$OQP is ${90}^{\circ }$

Through external angle theorem $\angle POR=\angle OQP+\angle OPQ$

$\Rightarrow {120}^{\circ }={90}^{\circ }+\angle OPQ$

$\angle OPQ={120}^{\circ }–{90}^{\circ }$

$\angle OPQ={30}^{\circ }$

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) ($\Rightarrow$ y=6-z)...   (2.1)

x+z = 7     …..(3)

Adding (1), (2) and (3), we have $2(x+y+z)=5+6+7=18$

$⟹x+y+z=\mathrm{9....}\left(4\right)$

Using (1) in (4), we have $5+z=9⟹z=4$

Now using, $\left(3\right)⟹x=7-z=7-4=3$

And $\left(2.1\right)⟹y=6-z=6-4=2$  

Therefore, the radii of the circles are 3 cm, 2 cm and 4 cm.

We know that

AQ = AR

AQ = AC + CR

But CR = CP

Therefore AQ = AC + CP    …..(i)

Also AQ = AB + BQ

But BQ = BP

$\therefore$ AQ = AB + BP     .….(ii)

Adding equations (i) and (ii)

2AQ = AC + CP + BP + AB

2AQ = Perimeter of ΔABC

Perimeter of $\Delta$ABC = 2 $\times$ 15 = 30 cm

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 $\perp$ AB
Let OA = a and OC = b.

Since OC $\perp$ AB, OC bisects AB     
[ $\because$ perpendicular from the centre to a chord bisects the chord].

In right $\Delta$ ACO, we have

$O{A}^2=O{C}^2+A{C}^2$    [by Pythagoras' theorem]

$\Rightarrow AC=\sqrt{O{A}^2-O{C}^2}=\sqrt{a^2-b^2}$
$\therefore AB=2AC=2\sqrt{a^2-b^2}$     [ $\because$ C is the midpoint of AB]

i.e., Length of the chord $AB=2\sqrt{a^2-b^2}$

AP = AS = 6 cm

BP = BA – AP = 11 – 6 = 5

BQ = BP = 5

CQ = CR = 4

x = CQ + BQ

x = 4 + 5 = 9 cm

ADB is a straight line. By linear pair axiom,
$\angle$ADP + $\angle$PDB = ${180}^{\circ }$
100 + $\angle$PDB = ${180}^{\circ }$
$\angle$PDB = ${80}^{\circ }$
Similarly $\angle$QED = ${80}^{\circ }$

We have, ∠DPE = ${90}^{\circ }$ (angle subtended by a diameter)

In $△DPE$,
$\angle$DPE + $\angle$PED + $\angle$EDP = 180
[Angle sum property of a triangle]
∠PED = ${10}^{\circ }$ 
Similarly $\angle$QDE = ${10}^{\circ }$

In $△DRE$,
$\angle$DRE + $\angle$RDE + $\angle$RED = 180
[Angle sum property of a triangle]
∠DRE = ${160}^{\circ }$

PRE is a straight line. By linear pair axiom,
$\angle$PRD + $\angle$DRE = ${180}^{\circ }$
$\angle$PRD + 160 = ${180}^{\circ }$
$\angle$PRD = ${20}^{\circ }$

We know that

$AD=AE$

$AD=AB+BE$

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), 

$AD+AD=AB+BF+CF+AC$

$2AD=AB+BC+AC$    

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.

$\therefore PA\perp OA,i.e.,\angle OAP={90}^{\circ }$      . .  . (i)
and  $PB\perp OB,i.e.,\angle OBP={90}^{\circ }$         . .. (ii)

Now, the sum of all the angles of a quadrilateral is ${360}^{\circ }$

$\therefore \angle AOB+\angle OAP+\angle APB+\angle OBP={360}^{\circ }$
$\Rightarrow \angle AOB+\angle APB={180}^{\circ }$  [using (i) and (ii)]

$\therefore$   quadrilateral OAPB is cyclic   [since both pairs of opposite angles have the sum ${180}^{\circ }$].

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.