Tag: tangents and intersecting chords

Questions Related to tangents and intersecting chords

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

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

  1. $p = q$

  2. $p^{2} = q^{2}$

  3. $q = -p$

  4. $p^{2} + q^{2} = 1$.

Show answer
Correct Option: A,B,C
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 $p^{2} =q^{2}$

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

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

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

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

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

  4. None of these.

Show answer
Correct Option: A
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.

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

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

  1. $h=r$

  2. $h=-r$

  3. ${r}^{2}+{h}^{2}=1$

  4. ${r}^{2}+{h}^{2}=2$

Show answer
Correct Option: A,B
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 $h=\pm r$

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 

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

  1. $4$

  2. $16$

  3. Does not exists 

  4. $8$

Show answer
Correct Option: C
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.

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

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

  1. $4$

  2. $3$

  3. $2$

  4. $1$

Show answer
Correct Option: C
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 $= \displaystyle \frac {2 \times 30}{5+12+13}=\frac {60}{30}=2cm$

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

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

  1. $16^{\circ}$

  2. $18^{\circ}$

  3. $27^{\circ}$

  4. $36^{\circ}$

Show answer
Correct Option: C
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
$\angle OAB + \angle OBA + \angle AOB = 180$
$2 \angle OAB + 126 = 180$
$\angle OAB = 27^{\circ}$

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

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

  1. 1 cm

  2. 2 cm

  3. 3 cm

  4. 4 cm

Show answer
Correct Option: B
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

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

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

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

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

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

  4. none

Show answer
Correct Option: C
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.

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.