Tag: other theorems related to circles

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Questions Related to other theorems related to circles

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

The range of values of $\lambda$ for which the circles $ { x }^{ 2 }+{ y }^{ 2 }=4$ and ${ x }^{ 2 }+{ y }^{ 2 }-2\lambda y+5=0$ have two common tangents only is-

  1. $\lambda \epsilon \left( -\sqrt { 5 } ,\sqrt { 5 } \right) $
  2. $\lambda <-\sqrt { 5 } or\quad \lambda >\sqrt { 5 }$
  3. $-\sqrt { 5 } <\lambda <1$

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

Two circles have two common tangents if the distance between their centers is less than the sum of their radii and greater than the difference of their radii. Here, centers are (0,0) and (0, lambda), radii are 2 and sqrt(lambda^2 - 5). The condition leads to the specified range.

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

Intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

  1. True

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

For any two parallel tangents to a circle, the segment of a third tangent intercepted between them subtends a 90-degree angle at the center because the radii to the points of tangency are perpendicular to the tangents.

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

$\overline { M N }$ and $\overline { M Q }$ are two tangents from a point $M$ to a circle with centre $0$ If $m \angle N O Q = 120 ^ { \circ } ,$ then ?

  1. $N Q = M N = M Q$
  2. $N Q = O M$
  3. $O Q = O M$
  4. $O N = M N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If angle NOQ = 120 degrees, then in the quadrilateral MONQ, the angles at N and Q are 90 degrees. Thus, angle M = 180 - 120 = 60 degrees. Triangle MNQ is isosceles with angle M = 60, so it is equilateral, meaning NQ = MN = MQ.

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

The chord of contact of the pair of tangents to the circle $x^2+y^2=1$ drawn from any point on the line $2x+y=4$ passes through a fixed point. 

  1. True

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

If chords are drawn to the circle from a fixed point $(x _1,y _1)$ and then tangents are drawn at point of contact, the point of intersection of all tangents lie on a fixed point.


The fixed point is called pole and fixed line is called polar.


Equation of polar is $T=0$.

$C:x^2+y^2-1=0$

Equation of polar is $T=0$.

$xx _1+yy _1-1=0$

The line is identical to given line $2x+y-4=0$.

By comparing coefficients, we get,
$\dfrac{x _1}{2}=\dfrac{y _1}{1}=\dfrac{-1}{-4}$

$x _1=\dfrac{1}{2},y _1=\dfrac{1}{4}$

Hence, the fixed point is $(\dfrac{1}{2}, \dfrac{1}{4})$.

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

From a point $P$ which is at a distance of $13$ cm from the centre $O$ of a circle of radius $5$ cm, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is:

  1. $60$ cm$^{2}$
  2. $65$ cm$^{2}$
  3. $30$ cm$^{2}$
  4. $32.5$ cm$^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The radius perpendicular tangent at the pt. of contact, therefore, $OQ\perp PQ$ and $OR\perp PR$
In rt. $\triangle OPQ$, we have
$PQ=\sqrt{OP^{2}-OQ^{2}}$
   $=\sqrt{169-25}=\sqrt{144}=12$ cm
$\Rightarrow $ $PR=12$ cm (Two tangents from the same external pt. to a circle are equal)
Now area of quad. $PQOR=2\times $Area of $\triangle POQ$
   $\displaystyle =\left ( 2\times \frac{1}{2}\times 12\times 5 \right )$ cm$^{2}=60$ cm$^{2}$

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

Circles ${ C } _{ 1 },{ C } _{ 2 },{ C } _{ 3 }$ have their centres at $\left( 0,0 \right) ,\left( 12,0 \right) ,\left( 24,0 \right) $ and have radii $1,2$ and $4$ respectively. Line ${t} _{1}$ is a common internal tangent to ${C} _{1}$ and ${C} _{2}$ and has a positive slope and line ${t} _{2}$ is a common internal tangent to ${C} _{2}$ and ${C} _{3}$ and has a negative slope. Given that lines ${t} _{1}$ and ${t} _{2}$ intersect at $(x,y)$ and that $x=p-q\surd r$, where $p,q$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r$.

  1. $p+q+r=26$
  2. $p+q+r=24$
  3. $p+q+r=28$
  4. $p+q+r=27$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

For the two circles ${ x }^{ 2 }+{ y }^{ 2 }=16$ and ${ x }^{ 2 }+{ y }^{ 2 }-2y=0$ there is/are

  1. One pair of common tangents

  2. Only one common tangent

  3. Three common tangents

  4. No common tangent
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The centres and radii of given circles are ${ C } _{ 1 }\left( 0,0 \right) ,{ r } _{ 1 }=4$ and ${ C } _{ 2 }\left( 0,1 \right) ,{ r } _{ 2 }=\sqrt { 0+1 } =1$
Now, ${ C } _{ 1 }{ C } _{ 2 }=\sqrt { 0+{ \left( 0-1 \right)  }^{ 2 } } =1$
and ${ r } _{ 1 }-{ r } _{ 2 }=4-1=3$
$\therefore { C } _{ 1 }{ C } _{ 2 }<{ r } _{ 1 }-{ r } _{ 2 }$
Hence, second circle lies inside the first circle.