Tag: tangents and intersecting chords

Questions Related to tangents and intersecting chords

$ABC$ is a right triangle with $\angle A = 90^{\circ}$. Let a circle touch tangent $\overline {AB}$ at A and tangent $\overline {BC}$ at some point D. Suppose the circle intersects $\overline {AC}$ again at E and $CE = 3 cm, CD = 6 cm$, find the measure of BD

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

  1. $9 cm$

  2. $3\sqrt {5} cm$

  3. $3 cm$

  4. $2 cm$

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Correct Option: A
Explanation:

Given $\angle BAC = 90^\circ$

$AB$ is tangent at $A$

$BDC$ is tangent at $D$

$CE = 3,CD =  6$

According to the tangent-secant theorem the length of tangent segment squared equals the product of secant segment and its external segment.

$\implies AC \times CE = CD^2$

$AC = \dfrac{6^2}{3} = 12$

$AE = 12 – CE = 9 = d = 2r$

Let $O$ be the center of the circle

$OA = OD = r$

$AB = BD = l$, tangents drawn from $B$

Since $\angle A = 90$

$BC^2 = AC^2 + AB^2$

$\implies (l + CD)^2 = AC^2 + l^2$

$\implies l^2 + 6^2 + 12l = 12^2 + l^2$

$\implies 12l = 108$

$BD = l = 9 \, cm = $ length of tangent

The value of $k$ for which two tangents can be drawn from $(k , k)$ to the circle $x^2 + y^2 + 2x + 2y 16 = 0$ is

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

  1. $k\ \epsilon\ R^+$

  2. $k\ \epsilon \ R$

  3. $k\ \epsilon\ ( -\infty , -4) \cup ( 2, \infty )$

  4. $k\ \epsilon\ ( 0, 1]$

Show answer
Correct Option: C
Explanation:

For two tangents to be drawn to $C(x, y):$

$\implies x^2 + y^2 +2x +2y – 16 = 0$

From P(k,k) , the point  P must lie outside the circle

$\implies C(k,k) > 0$

$\implies k^2 + k^2 + 2k + 2k – 16  > 0$

$\implies k^2 + 2k – 8 > 0$

$\implies (k + 4)(k - 2) > 0$

$\implies k \in (-\infty,-4) \cup (2,\infty)$

The area of the triangle formed by the tangents from the point $( 4, 3 )$ to the circle $x^{2} + y^{2} = 9$ and the line joining their points of contact is

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

  1. $\dfrac{25}{192}$ sq. units

  2. $\dfrac{192}{25}$ sq. units

  3. $\dfrac{384}{25}$ sq. units

  4. None of these

Show answer
Correct Option: B
Explanation:

Area of triangle $ = \dfrac {RL^3}{L^2 + R^2}$
where R = radius of the circle = 3
L = length of tangent $ = \sqrt {S _1} = \sqrt {16 + 9 - 9} = 4$
Hence area $ = \dfrac {192}{25} sq.$ units

The angle between the two tangents from the origin to the circle ${ \left( x-7 \right)  }^{ 2 }+{ \left( y+1 \right)  }^{ 2 }=25$ equals

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

  1. $\cfrac { \pi }{ 6 } $

  2. $\cfrac { \pi }{ 3 } $

  3. $\cfrac { \pi }{ 2 } $

  4. $\cfrac { \pi }{ 4 } $

Show answer
Correct Option: C
Explanation:

The equation of any line through the origin $(0,0)$ is

$y = mx +c$

If it is a tangent to the circle $(x−7)2+(y+1)2=52$, then

 

$ \dfrac{\left| 7m+1 \right|}{\left| \sqrt{{{m}^{2}}+1} \right|}=5 $

$ {{\left( 7m+1 \right)}^{2}}=5.\left( {{m}^{2}}+1 \right) $

$ 24{{m}^{2}}+14m-24=0 $

 

This equation, being a quadratic in m, gives two values of m, say ${{m} _{1}}$  and${{m} _{2}}$  These two values of m are the slopes of the tangents drawn from the origin to the given circle.

From (i), we have ${{m} _{1}}\times {{m} _{2}}=-1$

 

Hence, the two tangents are perpendicular.

 

Ans: C

For the circle ${ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }$, find the value of $r$ for which the area enclosed by the tangents drawn from the point $P(6,8)$ to the circle and the chord of contact is maximum.

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

  1. $5$

  2. $6$

  3. $8$

  4. $4$

Show answer
Correct Option: A
Explanation:

Let the angle between the tangents is $\theta\implies \tan \dfrac{\theta}{2}=\dfrac{r}{\sqrt{6^{2}+8^{2}-r^{2}}}$

$\implies \sin \dfrac{\theta}{2}=\dfrac{r}{10},\cos \dfrac{\theta}{2}=\dfrac{\sqrt{100-r^{2}}}{10}$
Area of triangle is $\dfrac{1}{2}(\sqrt{S _{11}})^{2}\sin \theta=\dfrac{r(100-r^{2})^{3/2}}{100}$
Let $f(x)=r(100-r^{2})^{3/2}\implies f'(x)=(100-r^{2})^{1/2}(100-4{r^{2})}=0\implies r^{2}=25\implies r=5$
For area to be maximum $r=5$

Write True or False and justify your answer in each of the following :


The length of tangent from an external point P on a circle with centre O is always less than OP.

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. Data insufficient

  4. Ambiguous

Show answer
Correct Option: B
Explanation:

$ Given-\ PA\quad &amp; \quad PB\quad are\quad tangents\quad to\quad the\quad circle\quad wtth\quad centre\quad O\ at\quad A\quad &amp; \quad B\quad respectively.\ To\quad find\quad out-\ The\quad assertion,\quad PA\quad or\quad PB\quad is\quad always\quad >\quad OA\quad or\quad OB,\quad is\quad \ true\quad or\quad false.\ Justification-\ PA\quad &amp; \quad PB\quad are\quad tangents\quad to\quad the\quad circle\quad at\quad A\quad &amp; \quad B\quad respectively.\ \therefore \quad PA\quad =\quad PB\quad and\quad \angle OAP\quad &amp; \quad \angle OBP\quad are={ 90 }^{ o }\ \therefore \quad \Delta OAP\quad is\quad a\quad right\quad one\quad with\quad \angle A={ 90 }^{ o }\ \Longrightarrow \angle AOP+\angle APO={ 90 }^{ o }\quad (by\quad angle\quad sum\quad prqperty\quad of\quad triangles)\ case\quad I-\quad \angle AOP=\angle APO\Longrightarrow each\quad of\quad them={ 45 }^{ o }.\quad i.e\quad PA=OA\ case\quad II-\quad \angle AOP>\angle APO\Longrightarrow \angle AOP>{ 45 }^{ o }\quad &amp; \quad \angle APO<{ 45 }^{ o }\quad \ (in\quad a\quad \Delta \quad the\quad side\quad opposite\quad to\quad the\quad greater\quad angle\quad is\quad greater\quad than\quad the\quad side\quad \ opposite\quad to\quad the\quad smaller\quad angle)\ \Longrightarrow PA>OA\ case\quad III-\quad \angle AOP<\angle APO\Longrightarrow \angle AOP{ <45 }^{ o }\quad &amp; \quad \angle APO>{ 45 }^{ o }\quad (\quad same\quad argument\quad as\quad case\quad II)\ \Longrightarrow PA>OA\ So\quad the\quad assertion,\quad PA\quad or\quad PB\quad is\quad always\quad <\quad OA\quad or\quad OB,\quad is\quad false.\ Ans-\quad False. $

To draw a pair of tangents to a circle which are inclined to each other at an angle of $60^0$, it is required to draw tangents at endpoints of those two radii of the circle, the angle between them should be

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

  1. $135^{0}$

  2. $90^{0}$

  3. $60^{0}$

  4. $120^{0}$

Show answer
Correct Option: D
Explanation:

Given:

$ PA$ & $PB$ are two tangents drawn from P to a circle with centre O at A & B respectively. 
$\angle APB={ 60 }^{ o }$ 
To find out-
$ \angle AOB$ 
Solution-
$ PA$ & $PB$ are two tangents drawn from P to the circle at A & B, respectively.
$ \therefore  PA=PB\Longrightarrow \Delta PAB$ is isosceles. 
i.e $\angle PAB=\angle PBA.$ 
or $\angle PAB+\angle PBA=2\angle PAB$  .....(i)
Now, $\angle PAB+\angle PBA+\angle APB={ 180 }^{ o }$    ....(angle sum property of triangles)
$ \Longrightarrow \angle PAB+\angle PBA{ +60 }^{ o }={ 180 }^{ O }$
$\Longrightarrow 2\angle PAB={ 120 }^{ o }$     ...(from i)
$ \therefore  \angle PAB={ 60 }^{ o }=\angle PBA$  .........(ii)
Again, $\angle OAP={ 90 }^{ o }$    ....(angle between a radius and the tangent at the point of contact.)
$ \therefore \angle OAB=\angle OAP-\angle PAB={ 90 }^{ o }-{ 60 }^{ o }$    .... (from ii)    .........(iii)
Since, $OA=OB$     ...(radii of the same circle)
$ \therefore  \Delta OAB$ is isosceles
$\Longrightarrow \angle OAB=\angle OBA={ 30 }^{ o }$   ....(from iii)
So, $\angle AOB={ 180 }^{ o }-(\angle OAB+\angle OBA)={ 180 }^{ o }-{ 30 }^{ o }-{ 30 }^{ O }={ 120 }^{ O }$      ...(angle sum property of triangles)

If two tangents inclined at an angle of $60^{\circ}$ are drawn to a circle of radius 3 cm, then length of the 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. $\sqrt{3}cm$

  2. $2\sqrt{3}cm$

  3. $\frac{2}{\sqrt{3}}cm$

  4. $3\sqrt{3}cm$

Show answer
Correct Option: D
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 D is the right answer.

The equation of tangent to the circle ${x^2} + {y^2} = 36$ which are incline at the angle of  ${45^ \circ }$ to the $x-$axis are 

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

  1. $x + y = \pm \sqrt 6 $

  2. $x = y \pm 3\sqrt 2 $

  3. $y = x \pm 6\sqrt 2 $

  4. None of these

Show answer
Correct Option: A

A tangent drawn from the point (4, 0) to the circle $\displaystyle x^{2}+y^{2}=8 $ touches it at a point A in the first quadrant. The coordinates of another point B on the circle such that $AB$ = 4 are

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

  1. $(2, -2)$

  2. $(-2, 2)$

  3. $\displaystyle \left ( -2\sqrt{2},0 \right ) $

  4. $\displaystyle \left ( 0,-2\sqrt{2} \right ) $

Show answer
Correct Option: A
Explanation:

$x^2  + y^2 = (2\sqrt 2)^2 , C = (0,0) , r = 2 \sqrt 2$

Let $y = mx + 2\sqrt 2 \sqrt{m^2 + 1}$ be a tangent

To find the tangent through $(4,0)$ substitute into the equation

$\implies 0 = 4m +2\sqrt 2 \sqrt{m^2 + 1}$

$\implies 16m^2 = 8(m^2 + 1)$

$\implies m = -1$

Equation is

$y = -x + 4$

Substituting in circle equation

$x^2 + (-x + 4)^2 = 8$

$\implies 2x^2 – 8x + 8 = 0$

$\implies x = 2 \implies y = 2$

$A = (2,2)$

Any point on the circle is given be$ (2\sqrt2 \cos \theta, 2\sqrt2 \sin \theta )$

Let B = $(2\sqrt2 \cos \theta _1, 2\sqrt2 \sin \theta _1)$

$AB = 4 \implies AB^2 = 16$

$\implies (2\sqrt2 \cos \theta _1, -2)^2 + (2\sqrt2 \sin \theta _1 - 2)^2 = 16$

$\implies 8 + 4 + 4 – 4\sqrt 2(\cos \theta _1 + \sin \theta _1) = 16$

$\implies \sin \theta _1 + \cos \theta _1 = 0$

$\theta _1 = \dfrac{-\pi}{4}$

$B = (2\sqrt 2 \times\dfrac{1}{\sqrt 2} 2\sqrt 2 \times\dfrac{-1}{\sqrt 2})  = (2,-2)$