Free Circles 03 Practice Test - 10th Grade 

The tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\therefore \angle OPC={90}^{\circ }$

$\text{Given, OP = PC.}$

$\text{So,}△OPC\text{is an isosceles right}\text{angled triangle.}\Rightarrow \angle PCO=\angle POC$

$\angle PCO+\angle POC+\angle OPC={180}^{\circ }\text{(Angle sum property of a triangle)}$

$\angle PCO+\angle POC+{90}^{\circ }={180}^{\circ }$

$\angle PCO+\angle POC={90}^{\circ }$

$\text{Hence,}\angle POC=\angle OCP={45}^{\circ }$

OMN  = ${90}^{\circ }$
Radius and tangent are perpendicular at point of contact remaining part of $\angle$OMN is $\angle OMP={90}^{\circ }-{60}^{\circ }={30}^{\circ }$
OP = OM = Radius
Hence, $\angle$OMP = $\angle$OPM = ${30}^{\circ }$

Therefore, $\angle$MOP = ${120}^{\circ }$.

Join the centre of the circle and the vertices of the triangle. Observe that the sides of the triangle become the tangents to the circle. Hence, the radii of the circle as shown in the question become the heights of the smaller triangles.

According to formula
$\frac{1}{2}$ (r) (sum of sides) = area of triangle
$\frac{1}{2}$ (r) ( 3+ 4 + 5) = $\frac{1}{2}$ × 3 × 4
r = 1 cm

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

Also, we know that if two tangents are drawn from an external point to a circle, then the line joining the external point and the centre of the circle bisects the angle between the tangents.

$⟹\angle APO=\angle OPB={60}^{\circ }$

Thus, cos$\angle OPA=cos{60}^{\circ }=\frac{AP}{OP}$ 
$⟹\frac{1}{2}$ = $\frac{AP}{OP}$

Thus, $OP=2AP$

To make tangents parallel to secant only two ways are possible.

Given that the area of circle = $25\pi c{m}^2$​
$⟹\pi r^2=25\pi$, where $r$ is the radius of the circle.
$i.e.,r^2=25⟹r=\pm 5$ 
Since radius is a non-negative quantity, we have $r=5cm$

Now, the distance between the two parallel lines
= Diameter of the circle
= $2\times$ Radius of the circle = 10 cm.

A parallelogram ABCD circumscribes a circle with centre O.

We know that the lengths of tangents drawn from an exterior point to a circle are equal.

$\therefore$    AP = AS,     . . . (i)    [tangents from A]

BP = BQ,                      . . . (ii)    [tangents from B]

CR = CQ,                     . . . (iii)    [tangents from C]

DR = DS                   . . . (iv)    [tangents from D]

Adding (i), (ii), (iii) and (iv), we get, 

AP + BP + CR + DR = AS + DS + BQ + CQ

AB + CD = AD + BC        . . . (v)

But we know that in a parallelogram opposite sides are equal.

$\therefore$  AD = BC and AB = CD, putting these in (v) we get 

2 AB = 2 AD

AB = AD = CD = BC.

Hence, ABCD is a rhombus.

Given that AO = AB=OB.

Since all sides are equal, $△AOB$ is equilateral, and hence equiangular. Also, each angle of the triangle equals ${60}^{\circ }.$
i.e.,$\angle$AOB =  ${60}^{\circ }$

 $⟹\angle ACB=\frac{1}{2}AOB$
($\because$ Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. )

$⟹\angle ACB=\frac{{60}^{\circ }}{2}={30}^{\circ }$

We know that the tangents drawn from an external point to a circle are equal.

$\therefore$    AD = AF,       . . . (i)     [tangents from A]

BD = BE,      . . . (ii)             [tangents from B]

CE = CF      . . . (iii)           [tangents from C]

Now, AB = AC         [given]

$\Rightarrow$         AD + BD = AF + CF

$\Rightarrow$         BD = CF

$\Rightarrow$    BE = CE             [using (ii) and (iii)]

$\Rightarrow$       E bisects BC.

$\angle$ABO= ${90}^{\circ }$ (point of contact)

Using Pythagoras theorem

${AO}^2$= ${AB}^2$+ ${BO}^2$

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

${25}^2$ - $7^2$ = ${AB}^2$

24= AB

length of the tangent = 24 cm