Tag: maths

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Questions Related to maths

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

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 

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

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Tangents to the circle x^2 + y^2 = r^2 inclined at an angle of 45 degrees to the x-axis have a slope of m = tan(45 degrees) = 1 or -1. Using the formula y = mx + r*sqrt(1 + m^2), with r = 6, we get y = x plus or minus 6 times square root of 2.

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

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

  1. $(2, -2)$
  2. $(-2, 2)$
  3. $\displaystyle \left ( -2\sqrt{2},0 \right ) $
  4. $\displaystyle \left ( 0,-2\sqrt{2} \right ) $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 = $(2\sqrt2 \cos \theta _1, 2\sqrt2 \sin \theta _1)$

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

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

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

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

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

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

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

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

  1. $\displaystyle \sqrt{\frac{bc}{a}}$
  2. $ac^2$
  3. $\displaystyle \frac{b}{a}$
  4. $\displaystyle \sqrt{\frac{c}{a}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $y=ax^2+bx+c$)

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

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 $R(5, 8)$ 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

  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. $x^2 + y^2 - 4x - 6y - 12 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circumcenter of triangle PQR, where P and Q are points of tangency from R, is the midpoint of the chord of contact and the center of the circle. Using the given circumcenter and radius, the equation can be derived.

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

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

  1. $\cfrac { 3 }{ \sqrt { 5 } } $
  2. $\cfrac { \sqrt { 5 } }{ 3 } $
  3. $\sqrt { 5 } $
  4. $\cfrac { 1 }{ \sqrt { 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

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

  1. 5

  2. 10

  3. 15

  4. 20

  5. 25
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $k=15$
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

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 = 

  1. 12 cm

  2. 5 cm

  3. 6 cm

  4. 4 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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

  1. $(1, 3)$
  2. $(3, 1)$
  3. $(1, 2)$
  4. $(2, 1)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $ 3 \, \pm \, \sqrt {60}$
  2. $ 4 \, \pm \, \sqrt {60}$
  3. $ 3 \, \pm \, \sqrt {6}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(3-\sqrt(60))$.

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

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

  1. $y=x$
  2. $y=-x$
  3. $y=1$
  4. $x=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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