Tag: maths

Questions Related to maths

If two tangents inclined at an angle $\displaystyle 60^{\circ}$ are drawn to a circle of radius 3 cm then 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. $\displaystyle \frac{3}{2}\sqrt{3}cm$

  2. $6 cm$

  3. $3 cm$

  4. $\displaystyle 3\sqrt{3}cm$

Show answer
Correct Option: D
Explanation:

Let PA and PB are the tangents on the circle. $\angle APB = 60$. the radius of the circle with center at O be 3 cm.
The two tangents drawn to a circle from an external point are equally inclined to the segment joining the center to the point.
Thus, $\angle APO = 30^{\circ}$
In $\triangle OAP$
$\angle OAP = 90^{\circ}$       ...(Angle between tangent and radius)
$\tan 30 = \cfrac{1}{\sqrt{3}} = \dfrac{OA}{AP}$
$PA = 3 \sqrt{3}$

Consider a curve $a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=1$ and a point $P$ not on the curve. A line drawn from the point $P$ intersect the curve ar point $Q$ and $R$. If the product $PQ.PR$ is independent of the slope of the line, then the curve is

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

  1. An ellipse

  2. A hyperbola

  3. A circle

  4. None of these

Show answer
Correct Option: C
Explanation:

Let the coordinates of point$P$ be $\left( { x } _{ 1 },{ y } _{ 1 } \right). $

Equation of any line through $P$ can be written as $\displaystyle \frac { x-{ x } _{ 1 } }{ \cos { \theta  }  } =\frac { y-{ y } _{ 1 } }{ \sin { \theta  }  } =r$    ...(1)
$\Rightarrow x={ x } _{ 1 }+r\cos { \theta  } ,y={ y } _{ 1 }+r\sin { \theta  } .$

Coordinates of any point an (1) is of the form $\left( { x } _{ 1 }+r\cos { \theta  } ,{ y } _{ 1 }+r\sin { \theta  }  \right) .$ 
This point will lie on ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }=1$ if
$a\left( { x } _{ 1 }+r\cos { \theta  }  \right) ^{ 2 }+2h\left( { x } _{ 1 }+r\cos { \theta  }  \right) \left( { y } _{ 1 }+r\sin { \theta  }  \right) +b{ \left( { y } _{ 1 }+r\sin { \theta  }  \right)  }^{ 2 }-1=0$
$\Rightarrow { r }^{ 2 }\left( a\cos ^{ 2 }{ \theta  } +2h\cos { \theta  } \sin { \theta  } +b\sin ^{ 2 }{ \theta  }  \right) +2\left[ { x } _{ 1 }\left( a\cos { \theta  } +h\sin { \theta  }  \right) +{ y } _{ 1 }\left( h\cos { \theta  } +b\sin { \theta  }  \right)  \right]$
$ +{ ax } _{ 1 }^{ 2 }+2{ hx } _{ 1 }{ y } _{ 1 }+{ by } _{ 1 }^{ 2 }-1=0$     ...(2)
Let $PQ={ r } _{ 1 }$  and $PR={ r } _{ 2 }.$ 
Then ${ r } _{ 1 },{ r } _{ 2 }$ are the roots of (2).
$\displaystyle \therefore PQ:PR={ r } _{ 1 }{ r } _{ 2 }=\frac { { ax } _{ 1 }^{ 2 }+2{ hx } _{ 1 }{ y } _{ 1 }+{ by } _{ 1 }^{ 2 }-1 }{ a\cos ^{ 2 }{ \theta  } +2h\cos { \theta  } \sin { \theta  } +b\sin ^{ 2 }{ \theta  }  } .$
We know rewrite the denominator.
We have$D=a\cos ^{ 2 }{ \theta  } +2h\cos { \theta  } \sin { \theta  } +b\sin ^{ 2 }{ \theta  } .\\ $
$\displaystyle =\frac { 1 }{ 2 } \left[ \left( a+b \right) +\left( a-b \right) \cos { 2\theta  }  \right] +h\sin { 2\theta  } $
$\displaystyle =\frac { a+b }{ 2 } +\frac { 1 }{ 2 } \left( a-b \right) \cos { 2\theta  } +h\sin { 2\theta  } $
Put $\displaystyle \frac { 1 }{ 2 } \left( a-b \right) =k\sin { \alpha  } ,h=k\cos { \alpha  } .$
$\displaystyle \Rightarrow k=\sqrt { { \left( \frac { a+b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 } } $ and $\displaystyle \tan { \alpha  } =\frac { a-b }{ 2h } $
$\displaystyle \therefore D=\frac { 1 }{ 2 } \left( a+b \right) +\sqrt { { \left( \frac { a-b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 } } \sin { \left( 2\theta +\alpha  \right)  } $
Thus, $\displaystyle PQ.PR=\frac { { ax } _{ 1 }^{ 2 }+2{ hx } _{ 1 }{ y } _{ 1 }+{ by } _{ 1 }^{ 2 }-1 }{ \frac { 1 }{ 2 } \left( a+b \right) +\sqrt { { \left( \frac { a-b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 } } \sin { \left( 2\theta +\alpha  \right)  }  } $
For  this to be independent of $\theta$ we must have $\displaystyle { \left( \frac { a-b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 }=0\Rightarrow a=b$ and $n=0.$
But this to be condition for the given curve to represent a circle.  

If $5x-12y+10=0$ and $12y-5x+16=0$ are two tangents
to a circle then radius of the circle is

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

  1. $1$

  2. $2$

  3. $4$

  4. $6$

Show answer
Correct Option: A
Explanation:

$5x-12y+10=0$ and $12y-5x+16=0$ are two parallel tangent to a circle.
Then distance $bet^{n}$ this two parallel tangents is $2r$.
$\therefore d=\left | \dfrac{-10-16}{\sqrt{5^{2}+12^{2}}} \right |=\left | \dfrac{26}{13} \right |=2$
$\therefore \ d=2r=2$
$\Rightarrow r=radius=1$

The equation to the locus of the point of intersection of any two perpendicular tangents to $x^{2}+ y^{2} = 4$ is

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

  1. $\mathrm{x}^{2}+\mathrm{y}^{2}=8$

  2. $\mathrm{x}^{2}+\mathrm{y}^{2}=12$

  3. $\mathrm{x}^{2}+\mathrm{y}^{2}=16$

  4. $\mathrm{x}^{2}+\mathrm{y}^{2}=4\sqrt{3}$

Show answer
Correct Option: A
Explanation:

The equation of the tangent to the circle $x^2+y^2=4$ is

$y=mx+2\sqrt{1+m^2}$
$P(h,k)$ lies on the tangent, then
$k-mh=2\sqrt{1+m^2}$
or, $(k-mh)^2=4(1+m^2)$
or, $m^2(h^2-4)-2mhk+k^2-4=0$
This is the quadratic equation in $m.$ Let $m _1$ and $m _2$ be roots
$m _1m _2=\cfrac{k^2-4}{h^2-4}=-1$
or, $k^2-4=-h^2+4$
or, $h^2+k^2=8$
Therefore, Equation to the locus of the intersection of any two perpendicular tangents is
$x^2+y^2=8$
Hence, A is the correct option.

If ${ \theta } _{ 1 },{ \theta } _{ 2 }$ be the inclinations of tangents drawn from the point $P$ to the circle ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ and $\cot { { \theta  } _{ 1 } } +\cot { { \theta  } _{ 2 } } =k$, then 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. $k\left( { y }^{ 2 }+{ a }^{ 2 } \right) =2xy$

  2. $k\left( { y }^{ 2 }-{ a }^{ 2 } \right) =2xy$

  3. $k\left( { y }^{ 2 }+{ a }^{ 2 } \right) =4xy$

  4. none of these

Show answer
Correct Option: B
Explanation:

Equation of the circle is ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$    ...(1)

Let $P$ be the point $\left( { x } _{ 1 },{ y } _{ 1 } \right) $.
Equation of any tangent to (1) is $y=mx+a\sqrt { 1+{ m }^{ 2 } } $
It is passes through $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $, then
${ y } _{ 1 }=m{ x } _{ 1 }+a\sqrt { 1+{ m }^{ 2 } } \Rightarrow { y } _{ 1 }-m{ x } _{ 1 }=a\sqrt { 1+{ m }^{ 2 } } $
Squaring ${ { y } _{ 1 } }^{ 2 }+2mx _{ 1 }{ y } _{ 1 }+{ m }^{ 2 }{ { x } _{ 1 } }^{ 2 }={ a }^{ 2 }\left( 1+{ m }^{ 2 } \right)$
$ \Rightarrow \left( { { x } _{ 1 } }^{ 2 }-{ a }^{ 2 } \right) { m }^{ 2 }-2{ x } _{ 1 }{ y } _{ 1 }m+\left( { { y } _{ 1 } }^{ 2 }-{ a }^{ 2 } \right) =0$   ...(2)
This is a quadratic in $m$. If ${ m } _{ 1 }$ and ${ m } _{ 2 }$ are its roots, then these are the slopes of the tangents from $P$.
Since inclination of tangents are given to be ${\theta} _{1}$ and ${\theta} _{2}$
$\therefore$ Let ${ m } _{ 1 }=\tan{{\theta} _{1}}$ and ${ m } _{ 2 }=\tan{{\theta} _{2}}$ 
$\displaystyle \Rightarrow \frac { 1 }{ { m } _{ 1 } } +\frac { 1 }{ { m } _{ 2 } } =k\Rightarrow { m } _{ 1 }+{ m } _{ 2 }=k{ m } _{ 1 }{ m } _{ 2 }$
$\displaystyle \therefore \frac { 2{ x } _{ 1 }{ y } _{ 1 } }{ { { x } _{ 1 } }^{ 2 }-{ a }^{ 2 } } =k.\frac { { { y } _{ 1 } }^{ 2 }-{ a }^{ 2 } }{ { { x } _{ 1 } }^{ 2 }-{ a }^{ 2 } } \Rightarrow 2{ x } _{ 1 }{ y } _{ 1 }=k\left( { { y } _{ 1 } }^{ 2 }-{ a }^{ 2 } \right) $
$\therefore $ Locus of $P$ is $k\left( { y }^{ 2 }{ -a }^{ 2 } \right) =2xy$

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

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

  1. $\displaystyle \frac{\pi}{3}$

  2. $\displaystyle \frac{\pi}{6}$

  3. $\displaystyle \frac{\pi}{2}$

  4. $\displaystyle \frac{\pi}{8}$

Show answer
Correct Option: C
Explanation:

$(x-7)^2+(y+1)^2=25$
PA=PB=length of tangent from $(0,0) \space  to \space  (x-7)^2+(y+1)^2-25=0$
$=\sqrt{51}$
$\Rightarrow PA=PB=\sqrt{7^2+1-25}=5$
In $\Delta  OAP,$
$\tan  \alpha =\dfrac{OA}{PA}=\dfrac{5}{5}=1$
$\alpha =45^{\circ}$
So, angle both tangents $ =2\alpha =90^{\circ}$

The number of tangents that can be drawn from (1, 2) to $x^2+y^2=5$ is

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

  1. 1

  2. 2

  3. 3

  4. 0

Show answer
Correct Option: A

Two secants PAB and PCD are drawn to a circle from an outside point P. Then, which of the following is true?

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

  1. PA. PB =PC +CD

  2. PA. PB =PC. PD

  3. PA+PB=PC+PD

  4. PA-PB = PC. CD

Show answer
Correct Option: B

State true or false
The angle between two tangents to circle may be ${0^0}$

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

true 

the angles between tangents to a circle can be zero if the tangents are parallel {if the tangents are drawn on opposite sides of a diameter}

State true or false
The length of tangent from an external point on a circle is always greater than the radius of the circle.

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

false, it is not always required it can even be less or greater than the radius of the circle, it depend on how far the point is from the center of the circle.