Tag: angle made by a chord and a tangent

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Questions Related to angle made by a chord and a tangent

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

The tangents drawn from the origin to the circle $x^{2} + y^{2} - 2px - 2qy + q^{2} = 0$ are perpendicular if

  1. $p = q$
  2. $p^{2} = q^{2}$
  3. $q = -p$
  4. $p^{2} + q^{2} = 1$.
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Equation of the given circle can be written as $(x -p)^{2} + (y -q)^{2} = p^{2}$
so, that the centre of the circle is $(p, q)$ and its radius is $p$.
This shows that $x = 0$ is a tangent to the circle from the origin.
Since tangents from the origin are perpendicular, the equation of the other tangent must be $y = 0$,
which is possible if $q = \pm  p $  or $p^{2} =q^{2}$

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 two tangents from the origin to the circle ${(x-7)}^{2}+{(y+1)}^{2}=25$ equals-

  1. $\cfrac{\pi}{2}$
  2. $\cfrac{\pi}{3}$
  3. $\cfrac{\pi}{4}$

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

Center is $(7,-1)$ and radius $=5$
Let equation of tangent from the origin be $y=mx$ $\Rightarrow mx-y=0$
Then, $\displaystyle\left| \frac { 7m+1 }{ \sqrt { { m }^{ 2 }+1 }  }  \right| =5$
$\Rightarrow { \left( 7m+1 \right)  }^{ 2 }=25\left( { m }^{ 2 }+1 \right) \Rightarrow 24{ m }^{ 2 }+14m-24=0$
Let ${ m } _{ 1 }$ and ${ m } _{ 2 }$ be the slopes of the two tangents.
Since $\displaystyle{ m } _{ 1 }{ m } _{ 2 }=-\frac{24}{24}=-1$
The two tangents are at right angles.

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

The tangents drawn from the origin to the circle ${ x }^{ 2 }+{ y }^{ 2 }-2rx-2hy+{h}^{2}=0$ are perpendicular if-

  1. $h=r$
  2. $h=-r$
  3. ${r}^{2}+{h}^{2}=1$
  4. ${r}^{2}+{h}^{2}=2$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Equation of the given circle can be written as ${ \left( x-r \right)  }^{ 2 }+{ \left( y-h \right)  }^{ 2 }={ p }^{ 2 }$
This has $(r,h)$ as the center and $r$ as the radius showing that it touches $y-$axis.
$\Rightarrow$ Other tangent from the origin to the circle must be $x-$axis which is possible if $h=\pm r$

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

If the tangents $PA$ and $PB$ are drawn from the point $P(-1,2)$ to the circle ${ x }^{ 2 }+{ y }^{ 2 }+x-2y-3=0$ and $C$ is the center of the circle, then the area of the quadrilateral $PACB$ is 

  1. $4$
  2. $16$

  3. Does not exists

  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given circle is $S:{ x }^{ 2 }+{ y }^{ 2 }+x-2y-3=0$

Since at point $P\left( -1,2 \right) $ ${ S } _{ \left( -1,2 \right)  }=1+4-1-4-3=-3<0$
the point $P(-1,2)$ lies inside the circle.
Consequently, the tangents from the point $P(-1,2)$ to the circle does not exits.
Thus, the quadrilateral $PACB$ cannot be formed.

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

In a right-angled triangle ABC, $\angle B=90^{o}, BC = 12 cm $ and $AB = 5 cm$.The radius of the circle inscribed in the triangle (in cm) is

  1. $4$
  2. $3$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know in $\triangle ABC, AB=5cm, BC=12cm$.
So, by pythagoras theorem we can find the length of side $AC$
$AC^2= AB^2 +BC^2=5^2+ 12^2$
$\therefore AC=13cm$
Circle is inscribed in a triangle. This type of circle is called as Incircle.
So, radius of incircle $=\displaystyle \frac {2 \triangle }{a+b+c}$
where $\triangle$ is the area of $\triangle ABC$ and $a,b,c$ are the sides of the triangle.
Area of $\triangle ABC= \displaystyle \frac {1}{2} AB \times BC= \frac {1}{2} \times 5 \times 12= 30sq.cm$
$\therefore$ radius of incircle $= \displaystyle \frac {2 \times 30}{5+12+13}=\frac {60}{30}=2cm$

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

In the given figure, if $PA$ and $PB$ are tangents to the circle with centre $O$ such that $\angle APB=54^{\circ},$ then $\angle OAB$ equals

  1. $16^{\circ}$
  2. $18^{\circ}$
  3. $27^{\circ}$
  4. $36^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $PA$ and $PB$ are the tangents from the point P.
$\angle APB = 54^{\circ}$
Now, In quadrilateral AOBP
$\angle OAP = \angle OBP = 90^{\circ}$ (Angle between tangent and radius)
Sum of angles = 360
$\angle OAP + \angle OBP + \angle OAB + \angle APB = 360$
$90 + 90 + 54 + \angle AOB = 360$
$\angle AOB = 126$

Now, In $\triangle OAB$
$OA = OB$ (Radius of circle)
$\angle OAB = \angle OBA$ (Isosceles triangle property)
Sum of angles = 180
$\angle OAB + \angle OBA + \angle AOB = 180$
$2 \angle OAB + 126 = 180$
$\angle OAB = 27^{\circ}$

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

ABC is a right angled triangle right angled at B such that $BC = 6$ cm and $AB = 8$ cm. A circle with center O is inscribed in $\displaystyle \Delta ABC$. The radius of the circle is

  1. 1 cm

  2. 2 cm

  3. 3 cm

  4. 4 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $BC = 6$ and $AB = 8$
using Pythagoras Theorem,
$AC^2 = AB^2 + BC^2$
$AC^2 = 6^2 + 8^2$
$AC = 10$
Radius = $\cfrac{2\times Area}{Perimeter}$
Radius = $\cfrac{2 \times (\dfrac{1}{2} \times 6 \times 8)}{10+8+6}$
Radius = $2$ cm

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

  1. $\displaystyle \frac{\pi }{4}$
  2. $\displaystyle \frac{\pi }{3}$
  3. $\displaystyle \frac{\pi }{2}$

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $y + 1 = m (x - 7) + \sqrt{25}(\sqrt{m^2 + 1})$ be any line to the circle.

Since we need tangents form $(0,0)$

$(0+1) = m(0 – 7) + 5\sqrt{m^2 + 1}$

$(7m + 1)^2 = 25(m^2 + 1)$

$\implies 24m^2 + 14m – 24 =0$

If $m _1, m _2$ are roots of the equation

$m _1m _2 = \dfrac{c}{a} = \dfrac{-24}{24} = -1$

Lines with $m _1$ and $m _2$ are slope are perpendicular.

Tangents from origin are at right angles to each other.

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.