Free Circles 01 Practice Test - 9th Grade 

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

$\therefore OP\perp OQ$

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

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

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

${OQ}^2=49+576$

${OQ}^2=625$

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

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

To find $\angle BAC,$

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

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

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

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

$\text{In}\Delta ABC,$

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

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

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
$\angle BOC=2\angle BAC=2\times {60}^{\circ }={120}^{\circ }.$

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
$\angle ADB+\angle DAB+\angle ABD={180}^{\circ }$

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

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

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.

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

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,
$\angle BCD+\angle DAB={180}^{\circ }.$
$⟹\angle BAD={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

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