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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

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

  1. $\displaystyle \frac{3}{2}\sqrt{3}cm$
  2. $6 cm$
  3. $3 cm$
  4. $\displaystyle 3\sqrt{3}cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
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)
$\tan 30 = \cfrac{1}{\sqrt{3}} = \dfrac{OA}{AP}$
$PA = 3 \sqrt{3}$

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

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

  1. An ellipse

  2. A hyperbola

  3. A circle

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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.  

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

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

  1. $1$
  2. $2$
  3. $4$
  4. $6$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $2r$.
$\therefore d=\left | \dfrac{-10-16}{\sqrt{5^{2}+12^{2}}} \right |=\left | \dfrac{26}{13} \right |=2$
$\therefore \ d=2r=2$
$\Rightarrow r=radius=1$

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

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

  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. $\mathrm{x}^{2}+\mathrm{y}^{2}=4\sqrt{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $k\left( { y }^{ 2 }{ -a }^{ 2 } \right) =2xy$

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

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

  1. $\displaystyle \frac{\pi}{3}$
  2. $\displaystyle \frac{\pi}{6}$
  3. $\displaystyle \frac{\pi}{2}$
  4. $\displaystyle \frac{\pi}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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}$

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

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

  1. PA. PB =PC +CD

  2. PA. PB =PC. PD

  3. PA+PB=PC+PD

  4. PA-PB = PC. CD
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By the power of a point theorem, for two secants PAB and PCD drawn from an external point P to a circle, the product of the segments of the secants is equal: PA * PB = PC * PD.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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.