Tag: maths

Questions Related to maths

The equation of tangent to the circle ${x^2} + {y^2} = 36$ which are incline at the angle of  ${45^ \circ }$ to the $x-$axis are 

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

  1. $x + y = \pm \sqrt 6 $

  2. $x = y \pm 3\sqrt 2 $

  3. $y = x \pm 6\sqrt 2 $

  4. None of these

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Correct Option: A

A tangent drawn from the point (4, 0) to the circle $\displaystyle x^{2}+y^{2}=8 $ touches it at a point A in the first quadrant. The coordinates of another point B on the circle such that $AB$ = 4 are

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

  1. $(2, -2)$

  2. $(-2, 2)$

  3. $\displaystyle \left ( -2\sqrt{2},0 \right ) $

  4. $\displaystyle \left ( 0,-2\sqrt{2} \right ) $

Show answer
Correct Option: A
Explanation:

$x^2  + y^2 = (2\sqrt 2)^2 , C = (0,0) , r = 2 \sqrt 2$

Let $y = mx + 2\sqrt 2 \sqrt{m^2 + 1}$ be a tangent

To find the tangent through $(4,0)$ substitute into the equation

$\implies 0 = 4m +2\sqrt 2 \sqrt{m^2 + 1}$

$\implies 16m^2 = 8(m^2 + 1)$

$\implies m = -1$

Equation is

$y = -x + 4$

Substituting in circle equation

$x^2 + (-x + 4)^2 = 8$

$\implies 2x^2 – 8x + 8 = 0$

$\implies x = 2 \implies y = 2$

$A = (2,2)$

Any point on the circle is given be$ (2\sqrt2 \cos \theta, 2\sqrt2 \sin \theta )$

Let B = $(2\sqrt2 \cos \theta _1, 2\sqrt2 \sin \theta _1)$

$AB = 4 \implies AB^2 = 16$

$\implies (2\sqrt2 \cos \theta _1, -2)^2 + (2\sqrt2 \sin \theta _1 - 2)^2 = 16$

$\implies 8 + 4 + 4 – 4\sqrt 2(\cos \theta _1 + \sin \theta _1) = 16$

$\implies \sin \theta _1 + \cos \theta _1 = 0$

$\theta _1 = \dfrac{-\pi}{4}$

$B = (2\sqrt 2 \times\dfrac{1}{\sqrt 2} 2\sqrt 2 \times\dfrac{-1}{\sqrt 2})  = (2,-2)$

A parabola $y = ax^2 + bx + c$ crosses the x-axis at $(\alpha, 0)$ $(\beta, 0)$ both to the right of the origin. A circle also passes through these two points. The length of the tangent from the origin to the circle is

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

  1. $\displaystyle \sqrt{\frac{bc}{a}}$

  2. $ac^2$

  3. $\displaystyle \frac{b}{a}$

  4. $\displaystyle \sqrt{\frac{c}{a}}$

Show answer
Correct Option: D
Explanation:

$OT$ is a tangent and $OAB$ is a secant 


we know that

$OT^2 =OA.OB$

         $=\alpha\beta$

         $=\dfrac{c}{a}$ (Since $\alpha,\beta $ are the roots of $y=ax^2+bx+c$)

$\Rightarrow OT=\sqrt{\dfrac{c}{a}}$

From a point $R(5, 8)$ two tangents $RP$ and $RQ$ are drawn to a given cirlce $S = 0$ whose radius is $5$. If circumcentre of the triangle PQR is $(2, 3)$, then the equation of circle $S= 0$ is

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

  1. $x^2 + y^2 + 2x + 4y - 20 = 0$

  2. $x^2 + y^2 + x + 2y - 10 = 0$

  3. $x^2 + y^2 - x - 2y - 20 = 0$

  4. $x^2 + y^2 - 4x - 6y - 12 = 0$

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Correct Option: A

The radius of the circle touching the straight lines $x-2y-1=0$ and $3x-6y+7=0$ is

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

  1. $\cfrac { 3 }{ \sqrt { 5 } } $

  2. $\cfrac { \sqrt { 5 } }{ 3 } $

  3. $\sqrt { 5 } $

  4. $\cfrac { 1 }{ \sqrt { 2 } } $

Show answer
Correct Option: B
Explanation:

Diameter of circle=distance of the point (1,0)
from $3x-6y+7=0$
$\therefore$ $\cfrac { 3(1)-6(0)+7 }{ \sqrt { { \left( 3 \right)  }^{ 2 }+{ \left( -6 \right)  }^{ 2 } }  } =\cfrac { 10 }{ \sqrt { 45 }  } =\cfrac { 2 }{ 3 } \sqrt { 5 } $
Now, radius of circle $=\cfrac { 1 }{ 2 } \left( \cfrac { 2 }{ 3 } \sqrt { 5 }  \right) =\cfrac { \sqrt { 5 }  }{ 3 } $

For what positive value(s) of K will the graph of the equation $2x + y = K$ be tangent to the graph of the equation $x^2+ y^2= 45$?

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

  1. 5

  2. 10

  3. 15

  4. 20

  5. 25

Show answer
Correct Option: C
Explanation:
  • The radius of circle is $\sqrt{45} = 3\sqrt5$ , center of circle is $(0,0)$
  • For the equation to be tangent to circle , the distance from center of circle to given line must be equal to radius of circle
  • So we get $k/\sqrt5 = 3\sqrt5$ , which gives $k=15$

AB and CD are two chords of a circle which when produced to meet at a point P such that AB = 5 cm, AP = 8 cm and CD = 2 cm then PD = 

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

  1. 12 cm

  2. 5 cm

  3. 6 cm

  4. 4 cm

Show answer
Correct Option: D
Explanation:

By intersecting secant theorem,

$PA$$\times$$PB$ = $PD$$\times$$PC$
$8$cm$\times$$3$cm = PD$\times$(PD+CD)
24${ cm }^{ 2 }$ = PD$\times$(PD+2)
${ PD }^{ 2 }$ $+ 2PD - 24 =0$
On Solving the above quadratic equation, we get
${ PD }^{ 2 }$$+6PD-4PD-24=0$
$(PD+6)$$\times$$(PD-4)=0$
$PD=4$cm & $-6$cm
So, $PD= 4$cm is the real solution

If the line $\displaystyle ax+by + c =0$ touches the circle $\displaystyle x^2 + y^2 -2x = \frac{3}{5}$ and is normal to the circle $\displaystyle x^2 + y^2 + 2x - 4y + 1 =0$, then $(a,b)$ are

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

  1. $(1, 3)$

  2. $(3, 1)$

  3. $(1, 2)$

  4. $(2, 1)$

Show answer
Correct Option: B
Explanation:

$x^2+y^2-2x=\dfrac {3}{5}\Rightarrow (x-1)^2+y^2=\dfrac {8}{5}$

So, Radius, $R=2\sqrt {\dfrac {2}{5}}$ and it's center is at $(1,0)$

ie, Distance, $d$ from the circle to $ax+by+c=0$ is,
$d=\dfrac {a\times 1+b\times 0+c}{\sqrt{a^2+b^2}}=\dfrac {a+c}{\sqrt{a^2+b^2}} =2\sqrt {\dfrac {2}{5}}\longrightarrow (1)$ (Inorder to satisfy the criterion of a tangent)

$x^2+y^2+2x-4y+1=0 \Rightarrow (x+1)^2+(y-2)^2=4$
So, It's center is at $((-1),2)$
As $ax+by+c=0$ is normal to the circle, it should go through the centre of the circle.
ie, $a-2b=c$ and $(y-2)=m(x+1)\longrightarrow (2)$

Substituting $c$ in (1),
$\dfrac {a+(a-2b)}{\sqrt{a^2+b^2}} =2\sqrt {\dfrac {2}{5}}$
$\Rightarrow \dfrac {a-b}{\sqrt {a^2+b^2}}=\sqrt {\dfrac {2}{5}}$

So, we can say $(a-b)=k\sqrt {2}$ and $a^2+b^2=5k^2$ foe some constant $k$.
$a^2+b^2-(a-b)^2=2ab=5k^2-2k^2=3k^2$
$(a-b)^2+4ab=(a+b)^2=6k^2+2k^2=8k^2\Rightarrow (a+b)=2k\sqrt{2}$
$a=\dfrac {1}{2}((a+b)+(a-b))=\dfrac {1}{2}(3k\sqrt{2})$
$b=\dfrac {1}{2}((a+b)-(a-b))=\dfrac {1}{2}(k\sqrt {2})$

Slope of the line, $m=\dfrac {dy}{dx}$
$\dfrac {d}{dx}(ax+by+c)=0\Rightarrow a+b\dfrac {dy}{dx}=0$
ie, $m=\dfrac {(-a)}{b}=(-3)$ (from above equations of $a$ and $b$)

Substituting the slope in (2),
$(y-2)=(-3)(x+1)\Rightarrow 3x+y+1=0$

Compairing with general equation given,
$(a,b)=(3,1)$

Option B is the correct answer.

Find $a$ if the distance between $(a , 2)$ and $(3 , 4)$  is $8 $

maths fundamentals pair of straight lines distances and midpoints distance formula in 2d

  1. $ 3 \, \pm \, \sqrt {60}$

  2. $ 4 \, \pm \, \sqrt {60}$

  3. $ 3 \, \pm \, \sqrt {6}$

  4. None of these

Show answer
Correct Option: A
Explanation:

By square of distance formual:


$8^2=(3-a)^2+(4-2)^2$

$=>64=a^2-6a+13$

$=>a^2-6a-51=0$

solving the quadratic we get:

$a=(3+\sqrt(60)$ or $(3-\sqrt(60))$.

The equation of the line which passes through $(0,0)$ and $(1,1)$ is ____________

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $y=x$

  2. $y=-x$

  3. $y=1$

  4. $x=1$

Show answer
Correct Option: A
Explanation:

The equation of the line which passes through (0,0) and (1,1) is y=x

As we know, the two points are $(0,0) (1,1)$
slope $ m$ $=\dfrac { { y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-{ x } _{ 1 } } $  $=\dfrac { 1-0 }{ 1-0 } =1$
Standard equation of the line is $y-{ y } _{ 1 }=m\left( x-{ x } _{ 1 } \right) $
Substituting value of $m =1 $ and point $(x,y) =  (0,0)$ we get:
$y-0=1\left( x-0 \right) $
 $y=x$ is the required equation of the line