Answer: Option D. -> 25 cm
:
D

Since tangent at a point on the circle is perpendicular to the radius through that point of contact.

$\therefore OP\perp OQ$

$\text{In right angled triangle OPQ,}$

${OQ}^2={OP}^2+{PQ}^2\text{(By Pythagoras theorem)}$

${OQ}^2=7^2+{24}^2$

${OQ}^2=49+576$

${OQ}^2=625$

$OQ=\sqrt{625}=25cm$

Answer: Option B. -> $(\frac{120}{3}{)}^{\circ }$
:
B and D

Given: ABCD is a cyclic quadrilateral whose side AB is the diameter of the circle and $\angle ADC={130}^{\circ }$.

To find $\angle BAC,$

 $\angle D+\angle B={180}^{\circ }\text{(Opposite angles of a cyclic quadilateral )}$

${130}^{\circ }+\angle B={180}^{\circ }$

$\angle B={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

$\angle ACB={90}^{\circ }\text{(Angle in a semicircle)}$

$\text{In}\Delta ABC,$

$\angle BAC+{50}^{\circ }+{90}^{\circ }={180}^{\circ }\text{(Since sum of angles of a triangle is}{180}^{\circ })$

$\angle BAC={180}^{\circ }-{90}^{\circ }-{50}^{\circ }={40}^{\circ }$

Answer: Option D. -> 120${}^{\circ }$
:
D

Given that $\Delta ABC$ is equilateral.
$⟹\angle BAC={60}^{\circ }$
Since the angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at any other point on the remaining part of the circle, we have
$\angle BOC=2\angle BAC=2\times {60}^{\circ }={120}^{\circ }.$

Answer: Option A. -> ${30}^{\circ }$
:
A

Given that ABCD is a cyclic quadilateral such that$\angle BCD={100}^{\circ }$ and $\angle ABD={70}^{\circ }.$

Since opposite angles of a cyclic quadrilateral are supplementary,
$\angle DAB+\angle DCB={180}^{\circ }.$

$⟹\angle DAB={180}^{\circ }-\angle DCB$
                       $={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Consider $△ADB$, since sum of angles of a triangle is ${180}^{\circ },$ we have
$\angle ADB+\angle DAB+\angle ABD={180}^{\circ }$

$i.e.,\angle ADB+{80}^{\circ }+{70}^{\circ }={180}^{\circ }$

$⟹\angle ADB={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

Answer: Option D. -> ${60}^{\circ }$
:
D

$\angle ADB={90}^{\circ }\text{[Angle in a semi-circle]}$
$\angle ABD=\angle ACD={30}^{\circ }\text{[Angles in the same segment]}$
In $△ADB$, we have
$\angle BAD+\angle ADB+\angle ABD={180}^{\circ }\text{(Sum of angles of a triangle is}{180}^{\circ })$
$x+{90}^{\circ }+{30}^{\circ }={180}^{\circ }$
$\Rightarrow x=({180}^{\circ }-{120}^{\circ })={60}^{\circ }$
$\text{Hence},x={60}^{\circ }$

Answer: Option D. -> ${60}^{\circ }$
:
Circle and line can interact in maximum three ways:

1. When a line is neither touching or cutting a circle.


2. When a line is touching a circle.


3. When a line is cutting a circle.

Answer: Option A. -> True
:
A
Draw an arc of smaller radius and join its ends at its centre. Now draw an arc of bigger radius and try to join its ends at its centre, now draw a straight line and try to join its ends at its centre we can see that it will meet its centre at infinity, so from this we can conclude that a line is a circle of infinite radius.

Take a triangle and add it to another triangle it will form a quadrilateral, add another triangle to this quadrilateral it will form a pentagon, keep adding triangles to this polygon at one point you will realize that a polygon with infinite sides will be a circle, with each side being infinitesimally small.

Answer: Option D. -> ${40}^{\circ }$
:
D

Since an angle in a semicircle is a right angle, $\angle ACB={90}^{\circ }.$
But, $\angle BCD=\angle ACD+\angle ACB={50}^{\circ }+{90}^{\circ }={140}^{\circ }.$
Consider the cyclic quadrilateral ABCD. Since opposite angles are supplementary,
$\angle BCD+\angle DAB={180}^{\circ }.$
$⟹\angle BAD={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

Answer: Option C. -> 13 cm
:
C

Given AB and CD are two chords of a circles on opposite sides of the centre. 
Construction: Draw perpendiculars OE and OF onto AB and CD respectively from centre O.
 AE = EB = 5cm and CF = FD = 12 cm
[ Perpendicular drawn to a chord from center bisects the chord]
Given,
Distance between two chords = 17 cm
Let distance between  O and F $=x$ cm
 And distance between  O and E  $=(17-x)cm$
In  ΔOEB,
$O{B}^2=O{E}^2+E{B}^2$ 
[Pythagoras theorem]
          $=(17-x{)}^2+5^2$ ---(1)
In  ΔOFD,
$OD2=O{F}^2+F{D}^2$
[Pythagoras theorem]
          $=(x{)}^2+{12}^2$----------→(2)
But OB = OD ( radii of the same circle).
From 1 & 2,
$(17-x{)}^2+5^2=(x{)}^2+{12}^2$
⇒ $289+x^2-34x+25=x^2+144$
⇒ $34x=170$
∴ $x=5$
Subsitute x in equation (2);
$O{D}^2=(5{)}^2+{12}^2=169$ 
$OD=13$
∴ Radius of the circle is 13 cm.

Answer: Option C. -> 10 cm
:
C

As the line segment BC passes through the center, it means that it is a diameter of the circle.
Given that radius of the circle = 5 cm
$\therefore$ Diameter = 5 + 5 = 10 cm