TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

Two circles of equal radius a cut orthogonally .If their centres are(2,3) and (5,6) then radical axis of these circles passes through the point 

Answer:

Explanation:

Let  $S_{1}$ be the circle with centre (2,3) and rradius a.

Let $S_{2}$ be the circle  with centre (5,6) and radius a

Then $S_{1}=(x-2)^{2}+(y-3)^{2}-a^{2}=0$

and $S_{2}=(x-5)^{2}+(y-6)^{2}-a^{2}=0$

 Now, radical axis of these circle is given by

$S_{1}-S_{2}$=0

  $(x-2)^{2}+(y-3)^{2}-a^{2}-(x-5)^{2}-(y-6)^{2}+a^{2}=0$

 $\Rightarrow$     $(4-4x+9-6y)-(25-10x)-(36-12y)=0$

 $\Rightarrow$  $13-4x-6y-25+10x-36+12y=0$

 $\Rightarrow$  $6x+6y-48=0$

 $\Rightarrow$    $x+y= \frac{48}{6}=8$

 $\Rightarrow$       $x+y=8$  ........(i)

 Since, the given  circle cut  othogonally, therefore

 $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$

$\Rightarrow$    $2(-2)(-5)+2(-3)(-6)=((-2)^{2}+(-3)^{2}-a^{2})+((-5)^{2}+(-6)^{2}-a^{2})$

  $[\because radius=\sqrt{g^{2}+f^{2}+c}\therefore a^{2}=g^{2}+f^{2}-c \Rightarrow c=g^{2}+f^{2}-a^{2}]$

$\Rightarrow$   $20+36=(13-a^{2})+(61-a^{2})$

$\Rightarrow$    $56=74-2a^{2}$

$\Rightarrow$    $2a^{2}=74-56=18$

$\Rightarrow$   $a^{2}=9$

$\Rightarrow$   a=3     $[ \because$ radius can't be negative ]

Clearly, option (c)  satisfy the Eq.(i).

In order to eliminate  the first degree terms from the equation

$4x^{2}+8xy+10y^{2}-8x-44y+14=0$ the point to which the origin has to be shifted is 

Answer:

Explanation:

On comparing the given equation with 

$ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ we get

 a=4,h=4, b=10,g=-4, f=-22 and c=14

 For removal of first degree terms , shift  the origin to

 $\left(\frac{bg-fh}{h^{2}-ab},\frac{af-gh}{h^{2}-ab}\right)=\left(\frac{-40+88}{16-40},\frac{-88+16}{16-40}\right)$

$=\left(\frac{48}{-24},\frac{-72}{-24}\right)=(-2,3)$

If the vectors   $a=\hat{i}+\hat{j}+\hat{k}$. $b=\hat{i}-\hat{j}+2\hat{k}$ and $c=x\hat{i}+(x-2)\hat{j}-\hat{k}$

are coplanar , then x=

Answer:

Explanation:

Given, 

$a=\hat{i}+\hat{j}+\hat{k}$.

$b=\hat{i}-\hat{j}+2\hat{k}$

and $c=x\hat{i}+(x-2)\hat{j}-\hat{k}$

 Since  , the given vectors are coplanar  , therefore

   [a  b  c]=0

  $\Rightarrow$    $\begin{bmatrix}1 & 1&1 \\1 & -1&2\\x&x-2&-1 \end{bmatrix}=0$

 $\Rightarrow$   $1(1-2x+4)-1(-1-2x)+1(x-2+x)=0$

 $\Rightarrow$   $5-2x+1+2x+2x-2=0$

 $\Rightarrow$   $2x+4=0$

 $\Rightarrow$    $2x=-4$

 $\Rightarrow$     $x=-2$

If rank of  $\begin{bmatrix}x & x&x \\x & x^{2}&x\\x&x&x+1 \end{bmatrix}$ is 1, then 

Answer:

Explanation:

A= $\begin{bmatrix}x & x&x \\x & x^{2}&x\\x&x&x+1 \end{bmatrix}$ 

Since , rank A=1 , therefore atleast  one determined of order 1 should be  non-zero  and all the determinants of order 2 and 3 should be zero.

 If x=0 , then $A=\begin{bmatrix}0 & 0&0 \\0 & 0&0\\0&0&1 \end{bmatrix}$ , which have non-zero

 determinant of order 1 only.

If x=1, then  $A=\begin{bmatrix}1 & 1&1 \\1 & 1&1\\1&1&2 \end{bmatrix}$, which have a non-zero determinant of order 2, 

nearly $\begin{bmatrix}1 & 1 \\1 & 2 \end{bmatrix}$

 $\therefore$  x can  take value 0 only

If $2x^{2}-10xy+2\lambda y^{2}+5x-16y-3=0$ represents a pair of straight  lines, then the point of intersection of those lines is 

Answer:

Explanation:

 On comparing  the given equations with

 $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ , we get

 $a=2, b=2 \lambda,h=-5, g=\frac{5}{2}$

     f=-8,c=-3

 Since , the given equations  represents a pair of straight  lines, therefore

$\begin{bmatrix}a & h&g \\h &b&f\\g&f&c \end{bmatrix}=0$

  $\Rightarrow\begin{bmatrix}2 & -5&5/2 \\-5 &2\lambda&-8\\5/2&-8&-3 \end{bmatrix}=0$

$\Rightarrow$      $2(-6 \lambda-64)+5(15+20)+\frac{5}{2}(40-5\lambda)=0$

$\Rightarrow$   $-12 \lambda-128+175+100-\frac{25}{2} \lambda=0$

$\Rightarrow$     $-\frac{49}{2} \lambda=-147$

$\Rightarrow$   $\lambda= \frac{147 \times 2}{49}=6$

Now, the point of intersection of given lines is given by

    $\left(\frac{bg-fh}{h^{2}-ab},\frac{af-gh}{h^{2}-ab}\right)=\left(\frac{30-40}{25-24},\frac{-16+\frac{25}{2}}{25-24}\right)$

              =$\left(-10,\frac{-7}{2}\right)$

• Mathematics - General questions