Tag: tangents and intersecting chords

Questions Related to tangents and intersecting chords

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


Correct Option: A

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

  1. $\left( -\sqrt { 5 } ,\sqrt { 5 } \right)$

  2. $\lambda <=\sqrt { 5 } or\quad \lambda >\sqrt { 5 }$

  3. $-\sqrt { 5 } <\lambda <1$

  4. none of these


Correct Option: A

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

  1. True

  2. False


Correct Option: A

In the given figure, $AD\ and AE$ are the tangents to a circle with centre $O\ and BC$ touches the circle at $F$. If $AE=5\ cm$ then perimeter of $\triangle ABC$ is 

  1. $15\ cm$

  2. $10\ cm$

  3. $22.5\ cm$

  4. $20\ cm$


Correct Option: A

$\overline { M N }$ 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. $O N = M N$


Correct Option: A

If $\triangle ABC$ is isoscles with $AB=AC$ and $C(O,r)$ is the incircle of the of the $\triangle BAC=30^{o}$. The tangent at $C$ intersects $AB$ at a point $D$, then $L$ trisects $BC$.

  1. True

  2. False


Correct Option: B

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


Correct Option: A
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 $(\dfrac{1}{2}, \dfrac{1}{4})$.

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$^{2}$


Correct Option: A
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$^{2}$

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. $p+q+r=27$


Correct Option: B

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


Correct Option: D
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.