Tag: maths

Questions Related to maths

If $t _{i}$ is the length of the tangent to the circle $ x^{2}+ y^{2} + 2g _{i} x + 5 =0; i =1,2,3$ from any point and $g _{1}, g _{2}$ and $g _{3} $ are in A.P. and $A _{i} = (g _{i},- t _{i}^{2})$, then

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

  1. $A _{1}, A _{2}, A _{3} $are collinear

  2. $A _{2}$ is the mid-point of $A _{1}$ and $A _{3} $

  3. $ A _{1} A _{2} $ is perpendicular. to $A _{2} A _{3}$

  4. $A _{2}$ divides $A _{1} A _{3}$ in the ratio $2: 5$

Show answer
Correct Option: A,B
Explanation:

$t _{i}^{2} = x^{2} + y^{2} + 2g _{i}x + 5$  where $(x, y) $ is any point. 


Since  $g _{1}, g _{2}, g _{3}$ are in $A.P.$

$\Rightarrow 2g _{2} = g _{1} + g _{3}$
$\Rightarrow  2t _{2}^{2}  = t _{1}^{2} + t _{3}^{2} \Rightarrow  t _{1}^{2},t _{2}^{2} ,t _{3}^{2} $ are in $A.P.$
and $A _{2}$  is the mid-point of $A _{1}$ and $A _{3}$.

$\Rightarrow  A _{1}, A _{2}, A _{3}$  are collinear.

If the area of the quadrilateral formed by the tangent from the origin to the circle $x^{2} +y^{2} +6x -10y

+ c = 0$ and the pair of radii at the points of contact of these tangents to tbe circle is $8$ square units. then $c$ is a root of the equation

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

  1. $ c^{2} -32c + 64 = 0$.

  2. $ c^{2} -34c + 64= 0$.

  3. $c^{2}+ 2c -64 = 0 $.

  4. $ c^{2} + 34c -64 = 0$.

Show answer
Correct Option: B
Explanation:

Let $OA, OB$ be the tangents from the origin to the given circle with centre $C(-3, 5)$
and radius .$\sqrt{9 + 25 -c}= \sqrt{ 34 -c} $
Then area of the quadribiteral $ OACB = 2 \times $ area of  $\triangle OAC = 2 \times (\dfrac 12) \times OA\times AC $
Now $OA =$ length of the tangent from the origin to the given circle $ = .\sqrt{C}$
and $AC =$ radius of the circle $=\sqrt{ 34 -c} $ so that. $\sqrt{C} \sqrt{34 -c} =8 $       ...(given)
$\Rightarrow  c (34 -c) =64 \Rightarrow c^{2} -34c+64=0$

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.