Tag: maths

Questions Related to maths

Circles ${ C } _{ 1 },{ C } _{ 2 },{ C } _{ 3 }$ have their centres at $\left( 0,0 \right) ,\left( 12,0 \right) ,\left( 24,0 \right) $ and have radii $1,2$ and $4$ respectively. Line ${t} _{1}$ is a common internal tangent to ${C} _{1}$ and ${C} _{2}$ and has a positive slope and line ${t} _{2}$ is a common internal tangent to ${C} _{2}$ and ${C} _{3}$ and has a negative slope. Given that lines ${t} _{1}$ and ${t} _{2}$ intersect at $(x,y)$ and that $x=p-q\surd r$, where $p,q$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r$.

  1. $p+q+r=26$

  2. $p+q+r=24$

  3. $p+q+r=28$

  4. $p+q+r=27$


Correct Option: B

For the two circles ${ x }^{ 2 }+{ y }^{ 2 }=16$ and ${ x }^{ 2 }+{ y }^{ 2 }-2y=0$ there is/are

  1. One pair of common tangents

  2. Only one common tangent

  3. Three common tangents

  4. No common tangent


Correct Option: D
Explanation:

The centres and radii of given circles are ${ C } _{ 1 }\left( 0,0 \right) ,{ r } _{ 1 }=4$ and ${ C } _{ 2 }\left( 0,1 \right) ,{ r } _{ 2 }=\sqrt { 0+1 } =1$
Now, ${ C } _{ 1 }{ C } _{ 2 }=\sqrt { 0+{ \left( 0-1 \right)  }^{ 2 } } =1$
and ${ r } _{ 1 }-{ r } _{ 2 }=4-1=3$
$\therefore { C } _{ 1 }{ C } _{ 2 }<{ r } _{ 1 }-{ r } _{ 2 }$
Hence, second circle lies inside the first circle.

From a point outside a circle, one tangent and one secant are drawn. The length of exterior part of secant is $7$ cm and that of interior part is $9$ cm. Find the length of tangent segment.

  1. $10.6$ cm

  2. $10.9$ cm

  3. $11.2$ cm

  4. $11.6$ cm


Correct Option: A
Explanation:

Let length of tangent be $l$

Length of exterior part of secant $=m=7 $ cm
Length of interior part of secant $=n=9 $ cm
Now using the secant intersection theorem, we have
${ l }^{ 2 }=m(m+n)\ \Rightarrow { l }^{ 2 }=7(7+9)\ \Rightarrow { l }^{ 2 }=112 $
$\Rightarrow l=\sqrt { 112 } =10.6$ cm
Option A is correct.

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is $60^0$. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

  1. 4 cm

  2. 6 cm

  3. 8 cm

  4. 10 cm


Correct Option: C
Explanation:

Join O and P.
Now, in triangles OPQ and OPR,
OP = OP (Common)
OQ = OR (radius of circle)
PQ = PR (tangents from single point)

Hence OPQ and OPR are congruent triangles.
$\angle OPQ = \angle OPR = 30^{\circ}$


Thus in triangle OPQ, $\dfrac{OQ}{OP} = Sin 30$

OP = $\dfrac{4}{Sin30}$

OP = 8 cm

From a point A which is at a distance of 10 cm from the center O of a circle of radius 6 cm, the pair of tangents AB and AC to the circle are drawn. Then the area of Quadrilateral ABOC is:

  1. $24 cm^{2}$

  2. $4 8cm^{2}$

  3. $96cm^{2}$

  4. $100cm^{2}$


Correct Option: B
Explanation:

Since $\triangle ABO$ is congruent to $\triangle ACO$,  area of $ABOC$ is twice the area of $\triangle ABO$.

In $\triangle ABO, \ OA = 10 cm, \ OB = 6 cm$.
Since tangent is perpendicular to radius at the point of contact, by Pythagoras' theorem, we have 
$AB = \sqrt{OA^2 - OB^2} = 8 cm$
So, the area of $\triangle ABO$ is $\dfrac{1}{2}\times AB\times OB = 24 cm^2$
So, the area of $ABOC$ is $2\times 24 = 48 cm^2$.
So option B is the right answer.

If the angle between two radii of a circle is $140^{\circ}$, then the angle between the tangents at the ends of the radii is :

  1. $90^{\circ}$

  2. $40^{\circ}$

  3. $70^{\circ}$

  4. $60^{\circ}$


Correct Option: B
Explanation:

Since tangents and radii are perpendicular at the point of contact, in the quadrilateral formed by the two radii and the tangents at their ends, we have two right angles at the two points of contacts.

Let the angle between the tangents be $x^o$. Then
$140 + 90 + 90 + x = 360 \Rightarrow x = 40^o$.
So option B is the right answer.

The lengths of tangents drawn from an external point to a circle are equal.

  1. True

  2. False

  3. Either

  4. Neither


Correct Option: A
Explanation:

Two tangents can be drawn from an external point and both their lengths are equal.

If two tangents inclined at an angle of $60^{\circ}$ are drawn to a circle of radius 3 cm, then the length of each tangent is equal to:

  1. $\dfrac{3\sqrt{3}}{2}$ cm

  2. $2\sqrt{3}$ cm

  3. $3\sqrt{3}$ cm

  4. 6 cm


Correct Option: C
Explanation:

Tangent is perpendicular to radius at the point of contact.

By symmetry with respect to the line joining the center and the point from which tangents are drawn, we have the length of tangent $=\dfrac{3}{\tan 30^o}=3\sqrt{3} cm$.
So option C is the right answer.

From point $P$ outside a circle, with a circumference of $10$ units, a tangent is drawn. Also from $P$ a secant is drawn dividing the circle into unequal arcs with lengths $m$ and $n$. It is found that $t$, the length of the tangent, is the mean proportional between $m$ and $n$. If $m$ and $t$ are integers, then $t$ may have the following number of values.

  1. Zero

  2. One

  3. Two

  4. Three


Correct Option: C
Explanation:

Circumference = 10 units

m+n=10
n=10-m
't' is the length of the tangent.
$t^{2}=mn$
$t=\sqrt{m(10-m)}$
At $m=1, t=3$
At $m=2, t=4$
$\therefore$ Two values are possible for t.

Tangents at the end points of  the diameter of a circle  intersect at angle Q Q is equal  to

  1. $90^{\circ}$

  2. $60^{\circ}$

  3. $0^{\circ}$

  4. $30^{\circ}$


Correct Option: C