Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

The locus of the centre of a circle touching the lines $x+2y=0$ and $x-2y=0$ is

  1. $xy=0$
  2. $x=0$
  3. $y=0$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

let $(h,k)$ be the center
Then distance from center to both lines will be equal
$\left| \dfrac { h+2k }{ \sqrt { 5 }  }  \right| =\left| \dfrac { h-2k }{ \sqrt { 5 }  }  \right| $
$\Rightarrow hk=0$

Ans: A

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

Consider a circle, $x^{2}+y^{2}=1$ and point $P\left(1,\sqrt{3}\right).PAB$ is secant drawn from $P$ intersecting circle in $A$ and $B$ (distinct) then range of $\left|PA\right|+\left|PB\right|$is 

  1. $\left[2\sqrt{3},4\right]$
  2. $\left(2\sqrt{3},4\right]$
  3. $\left(0,4\right]$
  4. $\left(0,2\sqrt{3}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Point P(1, sqrt(3)) lies on the circle x^2 + y^2 = 4 (radius 2). The secant PAB intersects the circle. The length PA + PB represents the sum of chord segments. The range is determined by the positions of the secant line.

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

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

  1. $2$
  2. $1$
  3. $0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(3,-2)$

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

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

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

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ${ x }^{ 2 }+{ y }^{ 2 }=0$)
$\Rightarrow { q }^{ 2 }={ p }^{ 2 }$

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

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

  1. A.P.

  2. G.P.

  3. H.P.

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 ${ x }^{ 2 }+{ y }^{ 2 }=2{ a }^{ 2 }$

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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

  1. $-3$
  2. $3$
  3. $-4$
  4. $4$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation
Required length
$\sqrt { { x }^{ 2 }+{ y }^{ 2 }+15x-17y+21-\left( { x }^{ 2 }+{ y }^{ 2 }+15x-17y+{ c }^{ 2 } \right)  } =\sqrt { 5 } $
$\Rightarrow \sqrt { 21-{ c }^{ 2 } } =\sqrt { 5 } \Rightarrow { c }^{ 2 }=16\Rightarrow c=\pm 4$
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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

  1. $12$
  2. $-1$
  3. $3$
  4. $11$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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) $

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

$\Rightarrow c=11,-1$
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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

  1. $9$
  2. $121$
  3. $ 205$
  4. $139$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$