Tag: tangents and intersecting chords

Quizzes (3) Questions (86)

Questions Related to tangents and intersecting chords

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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. $100cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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. $60^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A pair of tangents are drawn from a point $P$ to the circle $x^{2} + y^{2} = 1$. If the tangents make an intercept of $2$ on the line $x = 2$, the locus of $P$ is

  1. Straight line

  2. Pair of lines

  3. Circle

  4. Parabola
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $P$ be $(h, k)$ then by $SS _{1} = T^{2}$ the equation of pair of tangents drawn from $P$ to $x^{2} + y^{2} = 1$ is
$(h^{2} + k^{2} - 1)(x^{2} + y^{2} - 1) = (hx + ky - 1)^{2}$
Its intersection with the line $x = 1$ is given by
$y^{2}(h^{2} + k^{2} - 1) = (ky + h - 1)^{2}$
or $y^{2}(h^{2} - 1) - 2yk (h - 1) - (h - 1)^{2} = 0 .....(1)$
It is a quadratic in $y$ and we are given that length of intercept is $1 \therefore y _{1} - y _{2} = 2$
or $(y _{1} + y _{2})^{2} - 4y _{1}y _{2} = 1$
or $\left [\dfrac {2k(h - 1)}{h^{2} - 1}\right ]^{2} + 4\dfrac {(h - 1)^{2}}{h^{2} - 1} = 4$
or $\dfrac {4k^{2}}{(h + 1)^{2}} + \dfrac {4(h - 1)}{h + 1} = 4$
or $k^{2} = (h + 1)^{2} - (h^{2} - 1) = 2h + 2$
Hence the locus of $(h, k)$ is $y^{2} =2(x + 1)$ which represents a parabola.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A family of linear functions is given by $f(x) = 1 + c(x + 3)$ where $c \in R$. If a member of this family meets a unit circle centred at origin in two coincidence points then 'c' can be equal to

  1. $-3/4$
  2. $-1$
  3. $3/4$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given function,

$f(x)=1+c\left( x+3 \right)$                     .....( 1 )               where $c\in R$

We know that,

General equation of circle of radius $a$ meets at the origin is

${{x}^{2}}+{{y}^{2}}={{a}^{2}}$               

But it is unit circle then $a=1$,

Then, equation of circle is

$ {{x}^{2}}+{{y}^{2}}={{1}^{2}} $

$ {{x}^{2}}+{{y}^{2}}=1 $

Let  $f\left( x \right)=y$

By equation $\left( 1 \right)$ and we get,

$y=1+c\left( x+3 \right)$

$y=1+cx+3$                          ......( 2 )

Using distance formula,   at the origin to a line

$ d=\left| \dfrac{0-c\times 0-1-3c}{\sqrt{{{1}^{2}}+{{c}^{2}}}} \right|=1 $

$ \left| \dfrac{-1-3c}{\sqrt{{{1}^{2}}+{{c}^{2}}}} \right|=1 $

Taking square both side and we get, 

$ {{\left( \dfrac{1+3c}{\sqrt{1+{{c}^{2}}}} \right)}^{2}}=1 $

$ {{\left( 1+3c \right)}^{2}}=1+{{c}^{2}} $

$ {{1}^{2}}+{{\left( 3c \right)}^{2}}+6c=1+{{c}^{2}} $

$ 1+9{{c}^{2}}+6c=1+{{c}^{2}} $

$ 9{{c}^{2}}+6c-{{c}^{2}}=0 $

$ 8{{c}^{2}}+6c=0 $

$ 2c\left( 4c+3 \right)=0 $

$ 2c=0,4c+3=0 $

$ c=0,4c=-3 $

$ c=0,c=-\dfrac{3}{4} $

 Hence, It is required solution.