Tag: tangents and intersecting chords

Questions Related to tangents and intersecting chords

The number of tangents to the circle ${ x }^{ 2 }+{ y }^{ 2 }-8x-6y+9=0$ which passes through the point $(3,-2)$ is

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

  1. $2$

  2. $1$

  3. $0$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let $S\equiv { x }^{ 2 }+{ y }^{ 2 }-8x-6y+9=0$

Now $s$ for $(-3,2)=9+4-24+12+9>0$
$\therefore$ the point $(3,-2)$ lies outside the circle.
$\therefore$ $2$ tangents can be drawn to the circle from the point $(3,-2)$

Tangents drawn from the origin to the circle $ \displaystyle x^{2}+y^{2}-2px-2qy+q^{2}=0 $ are perpendicular to each other if

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

  1. $ \displaystyle p^{2}=q^{2} $

  2. $ \displaystyle p^{2}-q^{2}= 1 $

  3. $ \displaystyle p^{2}+q^{2}= 1 $

  4. None of these

Show answer
Correct Option: A
Explanation:

The equation of pair of tangents drawn from the origin to the given circle are $S{ S } _{ 1 }={ T }^{ 2 }$
$\Rightarrow \left( { x }^{ 2 }+{ y }^{ 2 }-2px-2qy+{ g }^{ 2 } \right) \left( 0+0-0-0+{ g }^{ 2 } \right) ={ \left( x.0+y.0-p\left( x+0 \right) -q\left( y+0 \right) +{ y }^{ 2 } \right)  }^{ 2 }$
$\Rightarrow { q }^{ 2 }\left( { x }^{ 2 }+{ y }^{ 2 }-2px-2qy+{ g }^{ 2 } \right) -{ \left( -px-qy+{ g }^{ 2 } \right)  }^{ 2 }=0$
The two tangents are $\bot $ if ${ g }^{ 2 }+{ q }^{ 2 }-{ p }^{ 2 }-{ g }^{ 2 }=0$
(Sum of coefficient of ${ x }^{ 2 }+{ y }^{ 2 }=0$)
$\Rightarrow { q }^{ 2 }={ p }^{ 2 }$

If the distance from the origin of the centers of the three circles ${ x }^{ 2 }+{ y }^{ 2 }+2{ a } _{ i }x={ a }^{ 2 }\left( i=1,2,3 \right) $ are in G.P., then the length of the tangent drawn to them from any point on the circle ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ are in

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

  1. A.P.

  2. G.P.

  3. H.P.

  4. none of these

Show answer
Correct Option: B
Explanation:

The centers of the three given circles are $\left( -{ \alpha  } _{ 1 },0 \right) ,\left( -{ \alpha  } _{ 2 },0 \right) $ and $\left( -{ \alpha  } _{ 3 },0 \right) $.

the distance of the three points from the origin are ${ \alpha  } _{ 1 },{ \alpha  } _{ 2 }$ and ${ \alpha  } _{ 3 }$.
Given: ${ \alpha  } _{ 1 },{ \alpha  } _{ 2 }$ and ${ \alpha  } _{ 3 }$ are in G.P.
$\Rightarrow { { \alpha  } _{ 2 } }^{ 2 }={ \alpha  } _{ 1 }{ \alpha  } _{ 2 }$
Now, coordinate of any point on the circle ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ are $\left( a\cos { \theta  } ,a\sin { \theta  }  \right) $.
$\therefore$ The lengths of the tangents drawn from the point $\left( a\cos { \theta  } ,a\sin { \theta  }  \right) $ to the three given circles are
$\sqrt { 2{ \alpha  } _{ 1 }a\cos { \theta  }  } ,\sqrt { 2{ \alpha  } _{ 2 }a\cos { \theta  }  } $ and $\sqrt { 2{ \alpha  } _{ 3 }a\cos { \theta  }  } $
using (1) are in G.P.

Two $ \displaystyle \perp $ tangents to the circle $ \displaystyle x^{2}+y^{2}=a^{2} $ meet at a point P. The locus of P has the equation

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

  1. $ \displaystyle x^{2}+y^{2}=3a^{2} $

  2. $ \displaystyle x^{2}+y^{2}=2a^{2} $

  3. $ \displaystyle x^{2}+y^{2}=4a^{2} $

  4. None of these

Show answer
Correct Option: B
Explanation:

The coordinates of $P$ be $(h,k)$. Then the equation of the tangents drawn from $P(h,k)$ to ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ is
$\left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) \left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 } \right) ={ \left( hx+hy-{ a }^{ 2 } \right)  }^{ 2 }$   (using SS'$={ T }^{ 2 }$)
This equation represents a pair of perpendicular lines.
Therefore, coefficient of ${ x }^{ 2 }$$+$ coefficient of ${ y }^{ 2 }=0$
$\Rightarrow \left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 }-{ h }^{ 2 } \right) +\left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 }-{ k }^{ 2 } \right) =0$
$\Rightarrow { h }^{ 2 }+{ k }^{ 2 }={ 2a }^{ 2 }$
Hence, locus is ${ x }^{ 2 }+{ y }^{ 2 }=2{ a }^{ 2 }$

The circle ${ x }^{ 2 }+{ y }^{ 2 }=4$ cuts the line joining the points $A(1,0)$ and $B(3,4)$ in two points P and Q. Let $\dfrac { BP }{ PA } =\alpha$ and $\dfrac { BQ }{ QA } =\beta$. Then $\alpha$ and $\beta$ are roots of the quadratic equation

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

  1. $3{ x }^{ 2 }+2x-21=0$

  2. $3{ x }^{ 2 }+2x+21=0$

  3. $2{ x }^{ 2 }+3x-21=0$

  4. None of these

Show answer
Correct Option: D
Explanation:
The equation of line joining $A(1,0)$ and $B(3,4)$ is $\dfrac{y-0}{x-1}=\dfrac{4-0}{3-1}\implies y=2x-2$    ... (1)

The point of intersection of this line and circle are

$x^2+(2x-2)^2=4\implies x^2+4x^2+4-8x=4\implies x=0,\dfrac{8}{5}$

Hence, points of intersection are $P(0,-2)$ and $Q\left(\dfrac{8}{5},\dfrac{6}{5}\right)$

Now, $BP=\sqrt{3^2+6^2}=\sqrt{45}$, $PA=\sqrt{1^2+2^2}=\sqrt{5}$, 

$BQ=\sqrt{\left(3-\dfrac{8}{5}\right)^2+\left(4-\dfrac{6}{5}\right)^2}=\sqrt{\dfrac{245}{25}}$ and $QA=\sqrt{\left(1-\dfrac{8}{5}\right)^2+\left(0-\dfrac{6}{5}\right)^2}=\sqrt{\dfrac{45}{25}}$

$\therefore \dfrac{BP}{PA}=3=\alpha$ and $\dfrac{BQ}{QA}=\dfrac{7}{3}=\beta$

The equation with roots $\alpha$ and $\beta$ is $(x-\alpha)(x-\beta)=0\implies (x-3)(3x-7)=0\implies 3x^2-16x+21=0$

This is the required answer.

If the length of the tangent drawn from any point on the circle $\displaystyle x^{2}+y^{2}+15x-17y+c^{2}=0$ to the circle $\displaystyle x^{2}+y^{2}+15x-17y+21=0 \ is \ \sqrt{5}$ units , then $c$ is equal to

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

  1. $-3$

  2. $3$

  3. $-4$

  4. $4$

Show answer
Correct Option: C,D
Explanation:
Required length
$\sqrt { { x }^{ 2 }+{ y }^{ 2 }+15x-17y+21-\left( { x }^{ 2 }+{ y }^{ 2 }+15x-17y+{ c }^{ 2 } \right)  } =\sqrt { 5 } $
$\Rightarrow \sqrt { 21-{ c }^{ 2 } } =\sqrt { 5 } \Rightarrow { c }^{ 2 }=16\Rightarrow c=\pm 4$

The area of the quadrilateral formed by the tangent from the point $(4, 5)$ to the circle $\displaystyle x^{2}+y^{2}-4x-2y-c=0$ with a pair of radii joining the points of contacts of these tangents is $8$ sq. units. The value of $c$ is

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

  1. $12$

  2. $-1$

  3. $3$

  4. $11$

Show answer
Correct Option: B,D
Explanation:

Given equation of circle is $x^2+y^2-4x-2y-c=0$


$(-g,-f)=(2,1)$

Radius $=\sqrt{g^2+f^2-c}$, $h$ of sub tangent 

Length of subtangent from point $(x _1,y _1) =\sqrt{x _1^2+y _1^2-4x _1-2y _1-c}$

 Area of quadrilateral = length of subtangent x radius

$\Rightarrow \sqrt { { 4 }^{ 2 }+{ 5 }^{ 2 }-4\times 4-2\times 5-c } \times \sqrt { 4+1+c } $

$ \Rightarrow { 8 }^{ 2 }=\left( 15-c \right) \left( 5+c \right) $

$\Rightarrow { c }^{ 2 }-10c-11=0$

$\Rightarrow c=11,-1$

A line is drawn through the point $P(3, 11)$ to cut the circle $x^{2}+y^{2}= 9$ at $A$ and $B$. Then $PA\cdot PB$ is equal to

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

  1. $9$

  2. $121$

  3. $ 205$

  4. $139$

Show answer
Correct Option: B
Explanation:

From geometry we know $PA\cdot  PB = (PT)^{2}$

where $PT$ is the length of the tangent from $P$ to the circle.

Hence $PA\cdot PB=

(3)^2 + (11)^{2} - 9 = 11^{2} = 121$

If $t _{i}$ is the length of the tangent to the circle $ x^{2}+ y^{2} + 2g _{i} x + 5 =0; i =1,2,3$ from any point and $g _{1}, g _{2}$ and $g _{3} $ are in A.P. and $A _{i} = (g _{i},- t _{i}^{2})$, then

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

  1. $A _{1}, A _{2}, A _{3} $are collinear

  2. $A _{2}$ is the mid-point of $A _{1}$ and $A _{3} $

  3. $ A _{1} A _{2} $ is perpendicular. to $A _{2} A _{3}$

  4. $A _{2}$ divides $A _{1} A _{3}$ in the ratio $2: 5$

Show answer
Correct Option: A,B
Explanation:

$t _{i}^{2} = x^{2} + y^{2} + 2g _{i}x + 5$  where $(x, y) $ is any point. 


Since  $g _{1}, g _{2}, g _{3}$ are in $A.P.$

$\Rightarrow 2g _{2} = g _{1} + g _{3}$
$\Rightarrow  2t _{2}^{2}  = t _{1}^{2} + t _{3}^{2} \Rightarrow  t _{1}^{2},t _{2}^{2} ,t _{3}^{2} $ are in $A.P.$
and $A _{2}$  is the mid-point of $A _{1}$ and $A _{3}$.

$\Rightarrow  A _{1}, A _{2}, A _{3}$  are collinear.

If the area of the quadrilateral formed by the tangent from the origin to the circle $x^{2} +y^{2} +6x -10y

+ c = 0$ and the pair of radii at the points of contact of these tangents to tbe circle is $8$ square units. then $c$ is a root of the equation

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

  1. $ c^{2} -32c + 64 = 0$.

  2. $ c^{2} -34c + 64= 0$.

  3. $c^{2}+ 2c -64 = 0 $.

  4. $ c^{2} + 34c -64 = 0$.

Show answer
Correct Option: B
Explanation:

Let $OA, OB$ be the tangents from the origin to the given circle with centre $C(-3, 5)$
and radius .$\sqrt{9 + 25 -c}= \sqrt{ 34 -c} $
Then area of the quadribiteral $ OACB = 2 \times $ area of  $\triangle OAC = 2 \times (\dfrac 12) \times OA\times AC $
Now $OA =$ length of the tangent from the origin to the given circle $ = .\sqrt{C}$
and $AC =$ radius of the circle $=\sqrt{ 34 -c} $ so that. $\sqrt{C} \sqrt{34 -c} =8 $       ...(given)
$\Rightarrow  c (34 -c) =64 \Rightarrow c^{2} -34c+64=0$