Tag: tangents and intersecting chords

Questions Related to tangents and intersecting chords

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.

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

  1. $10.6$ cm

  2. $10.9$ cm

  3. $11.2$ cm

  4. $11.6$ cm

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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.

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

  1. 4 cm

  2. 6 cm

  3. 8 cm

  4. 10 cm

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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:

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

  1. $24 cm^{2}$

  2. $4 8cm^{2}$

  3. $96cm^{2}$

  4. $100cm^{2}$

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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 :

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

  1. $90^{\circ}$

  2. $40^{\circ}$

  3. $70^{\circ}$

  4. $60^{\circ}$

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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.

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

  1. True

  2. False

  3. Either

  4. Neither

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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:

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

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

  2. $2\sqrt{3}$ cm

  3. $3\sqrt{3}$ cm

  4. 6 cm

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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.

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

  1. Zero

  2. One

  3. Two

  4. Three

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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

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

  1. $90^{\circ}$

  2. $60^{\circ}$

  3. $0^{\circ}$

  4. $30^{\circ}$

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Correct Option: C

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

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

  1. Straight line

  2. Pair of lines

  3. Circle

  4. Parabola

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Correct Option: D
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.

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

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

  1. $-3/4$

  2. $-1$

  3. $3/4$

  4. $1$

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Correct Option: A
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.