Free Circles 01 Practice Test - 10th Grade 

Given that OP = 25 cm and OQ = 7 cm.

To find the length of PQ, apply Pythagoras theorem in $△$OPQ since $\angle OQP={90}^{\circ }$

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

$7^2+{QP}^2={25}^2$

${QP}^2=625–49=576$

$QP=24cm$
The length of the tangent is 24 cm.

Given that $\angle POQ={110}^{\circ }$
Note that $\angle OQT=\angle OPT={90}^{\circ }$
($\because$ A tangent at any point of a circle is perpendicular to the radius at the point of contact)

Also, $\angle TQO+\angle QOP+\angle OPT+\angle PTQ={360}^{\circ }$($\because$ Sum of interior angles of a quadrilateral is 360 degrees)

$⟹\angle PTQ={360}^{\circ }–{90}^{\circ }–{90}^{\circ }–{110}^{\circ }$

$⟹\angle PTQ={70}^{\circ }$

$\therefore \angle PTQ={70}^{\circ }$

Given that AO = 10 cm, BO = 6 cm

Applying Pythagoras theorem

${AB}^2={AO}^2–{BO}^2$

${AB}^2={10}^2–6^2$

$\Rightarrow {AB}^2=100–36=64$

$AB=8$ cm

At most one circle can be drawn through a given set of three distinct points. These three points will then be referred to as 'concyclic points' (Lying on the same circle). .

Take any secant like AB as shown, there cannot be more than 2 tangents parallel to it. So, the answer is 2 not 4.

Since,
tangent to  a circle is perpendicular to the radius through the point of contact 
So, $\angle OTP={90}^0$
So, in triangle OTP
$(OP{)}^2=(OT{)}^2+(PT{)}^2$
${13}^2=(OT{)}^2+{12}^2$
$(OT{)}^2={13}^2-{12}^2$
$O{T}^2=25$
$OT=\sqrt{25}$
$OT=5$
So, radius of the circle is 5 cm

We have

OA $\perp$ AP and OB $\perp$ BP [ The tangent at any point of a circle is perpendicular to the radius through the point of contact].

Join OP.

In right $\Delta$ OAP, we have

OA = 8 cm, AP = 6 cm

$\therefore O{P}^2=O{A}^2+A{P}^2$  [by Pythagoras theorem]

$\Rightarrow OP=\sqrt{O{A}^2+A{P}^2}=\sqrt{8^2+6^2}cm=\sqrt{100}cm=10cm$

In right $\Delta$ OBP, we have

OB = 6 cm, OP = 10 cm

$\therefore O{P}^2=O{B}^2+B{P}^2$
  [by Pythagoras' theorem]

$\Rightarrow BP=\sqrt{O{P}^2-O{B}^2}=\sqrt{{10}^2-6^2}cm=\sqrt{64}cm$

Thus, the length of BP
 $=\sqrt{64}cm$ = 8 cm.

Given that AC = BD.

OA = OB (radius)

Adding the above equations,

AC + OA = BD + OB

$⟹$OC = OD

$⟹\angle ODC=\angle OCD$ (Property of isosceles triangles)

Given $\angle COD={90}^{\circ }$

Therefore, $\angle ODC=\angle OCD={45}^{\circ }$ (Base angles are equal)

Join OP.

Since a tangent at any point of a circle is perpendicular to the radius at the point of contact, we have OP $\perp$ CD.

Consider right angled triangle ODP,
 $tan\angle ODP=\frac{OP}{PD}$

$Tan{45}^{\circ }=\frac{100}{PD}(\because OP=1m=100cm)$

$⟹1=\frac{100}{PD}$

$\Rightarrow PD=100\text{cm}$
Similarly we can show that PC = 100 cm

Now, $sin\angle ODP=\frac{OP}{OD}$

$⟹sin{45}^{\circ }=0.7071=\frac{100}{OD}$

i.e., $OD=\frac{100}{0.7071}$

$\Rightarrow OD=141.42\text{cm}$

But, $BD=OD-OB$

$\Rightarrow BD=141.42–100=41.42\text{cm}$

$AC+CP=BD+100$ (since AC = BD and PC= 100 cm)
$=41.42+100=141.42cm$

A line that touches a circle at only one point is called a tangent.

In $\Delta AOC$,
OA=OC      --------(radii of the same circle)

$\therefore \Delta AOC$ is an isosceles triangle
$\to \angle OAC=\angle OCA={25}^{\circ }$----- (base angles of an isosceles triangle )

In $\Delta BOC$,
OB=OC      --------(radii of the same circle)
$\therefore \Delta BOC$ is an isosceles triangle
$\to \angle OBC=\angle OCB={35}^{\circ }$ -----(base angles of an isosceles triangle )

$\angle ACB={25}^{\circ }+{35}^{\circ }={60}^{\circ }$
$\angle AOB=2\times \angle ACB$ ----(angle at the center is twice the angle at the circumference)
           
             = 2$\times {60}^{\circ }$
             =${120}^{\circ }$