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 The number of common tangents to  the circles   $x^{2}+y^{2}-4x-6y-12=0$  and $x^{2}+y^{2}+6x+18y+26 =0$  is

Answer:

Explanation:

Central Idea:

  Number of common tangents depend on the position of the circle with respect to each other.

  (i) If circles touch externally   $\Rightarrow$    C₁ C₂  =r₁+r₂   , 3 common tangents

  (ii)  If circles touch internally  $\Rightarrow$   C₁ C₂  =r₂ -r₁ , 1 common integer

   (iii) If circles do not touch each other, 4 common tangents

 Given equations of circles are

      $x^{2}+y^{2}-4x-6y-12=0$      ......(i)

      $x^{2}+y^{2}+6x+18y+26 =0$     .......(ii)

  Centre of circle (i) C₁   (2,3) and radius 

                           = $\sqrt{4+9+12}=5(r_{1})$    (say)

 Centre of circle (ii) is C₂  (-3,-9)  and radius

                     =$\sqrt{9+81-26}=8(r_{2})$               (say)

Now, C₁ C₂=   $\sqrt{(2+3)^{2}+(3+9)^{2}}$

$\Rightarrow$     $C_{1}C_{2}=\sqrt{5^{2}+12^{2}}$

  $\Rightarrow$    $C_{1}C_{2}=\sqrt{25+144}=13$

 $\therefore$      $r_{1}+r_{2}=5+8=13$

  Also, C₁ C₂  = r₁ +r₂

  Thus, both circles touch each other externally. Hence there are three common tangents

 Locus of the image of the point (2,3) in the line (2x-3y+4)+k(x-2y+3)=0 k ε  R  is a

Answer:

Explanation:

Central Idea      First of all find the point of intersection of the lines 2x-3y+4=0 and x-2y+3=0 (say   A) . Now  , the line

(2x-3y+4) +k(x-2y+3)=0 is the perpendicular bisector of the line joining points P(2,3) and image P ' (h,k). Now  AP= AP '  and simplify

  Given line is

   (2x-3y+4)+k(x-2y+3)=0 , k ε   R  ..... (i)

   This line will pass through the point of intersection of the lines

                  2x-3y+4= 0       .......(ii)

 and          x-2y+3= 0      ..........(iii)

 On solving Eqs (ii) and (iii), we get

  x=1, y=2

  $\therefore$ Point of intersection of lines (ii) and (iii)  is (1,2)

Let M be the mid-point  of PP' , then AM is perpendicular bisector of PP' (where A, is the point of intersection of given lines )

  $\therefore$       AP=AP'

   $\Rightarrow$     $\sqrt{(2-1)^{2}+(3-2)^{2}}=\sqrt{(h-1)^{2}+(k-2)^{2}}$

 $\Rightarrow$      $\sqrt{2}=\sqrt{h^{2}+k^{2}-2h-4k+1+4}$

 $\Rightarrow$      $\sqrt{2}=\sqrt{h^{2}+k^{2}-2h-4k+5}$

 $\Rightarrow$     $\sqrt{2}=\sqrt{h^{2}+k^{2}-2h-4k+5}$

 $\Rightarrow$    $h^{2}+k^{2}-2h-4k+3=0$

Thus  , the required locus is

              $x^{2}+y^{2}-2x-4y+3=0$

  which is a equation  of circle with

    radius  =   $\sqrt{1+4-3}=\sqrt{2}$

  Aliter

  (2x-3y+4)+k(x-2y+3)=0     is family of line passing through (1,2)  By congruency of triangles . We can prove that mirror image (h,k) and the point (2,3) will be equidistant  from (1,2)

                         $\therefore$  Locus of (h,k)  PR=PQ

             $\Rightarrow  (h-1)^{2}+(k-2)^{2}= (2-1)^{2}+(3-2)^{2}$

or       $ (x-1)^{2}+(y-2)^{2}= 2$

  $\therefore$     Locus is a circle of radius =   $\sqrt{2}$

The number of points having both coordinates as integers that lie in the interior of the triangle with vertices (0,0),(0,41) and (41,0). is

Answer:

Explanation:

Required points (x,y) are such that is satisfy  x+y <41

   and    x>0,y>0

   Number of positive integral solution of the  equation x+y+k=41 will be number of integral coordinates in the bounded region.

  $\therefore$   Total number of integral  coordinates

        $=^{41-1}C_{3-1}=^{40}C_{2}=\frac{40!}{2!38!}=780$

  Alter

 Consider the following figure.

Clearly, the number of requires points

            = 1+2+3+.......+39

    $=\frac{39}{2}(39+1)=780$

Let y(x) be the solution of the differential equation  $(x \log x) \frac{\text{d}y}{\text{d}x}+y=2x\log x. (x\geq1) $ . Then , y(e) is equal to

Answer:

Explanation:

 Given differentail equation is

          $(x\log x)\frac{\text{d}y}{\text{d}x}+y=2x \log x,$

    $\Rightarrow $    $\frac{\text{d}y}{\text{d}x}+\frac{y}{x\log x}=2$

 This is a linear  differential  equation.

    $\therefore$     $IF=e^{\int_{}^{} \frac{1}{x log x}dx}=e^{\log(\log x)}=\log x$

 Now,   the solution of given differential equation is given by

    $y. \log x=\int_{}^{} \log x.2dx$

     $\Rightarrow y. \log x=2\int_{}^{} \log x.dx$

  $\Rightarrow y. \log x=2[x \log x-x]+c$

  At   X=1,  c=2

  $\Rightarrow$     $y. \log x=2[x \log x-x]+2$

   At           x=e,

      y= 2(e-e)+2   $\Rightarrow y=2$

The area (in sq units) of the  region described by   $(x,y:y^{2}\leq2x)$ and  $(y\geq4x-1)$  is 

Answer:

Explanation:

 Given region is       $(x,y:y^{2}\leq2x)$ and  $(y\geq4x-1)$                          

    $y^{2}\leq 2x$   represents  a region inside the  parabola

           $y^{2}= 2x$                ..............(i)

 and  $y\geq 4x-1$   represents   a region to the left of the line

                           y=4x-1                    .......(ii)

 The point of intersection of the curve (i) and (ii) is 

                      $( 4x-1)^{2}=2x$

$\Rightarrow$       $16x^{2}+1-8x=2x$

$\Rightarrow$       $16x^{2}-10x+1=0$

$\Rightarrow$      $x=\frac{1}{2},\frac{1}{8}$

$\therefore$    The points where these curves interset , are    $\left(\frac{1}{2},1\right)$  and   $\left(\frac{1}{8},\frac{1}{2}\right)$

Hence, required area

                    =  $\int_{-1/2}^{1} \left( \frac{y+1}{4}-\frac{y^{2}}{2}\right)dy$

                 =   $\frac{1}{4}\left(\frac{y^{2}}{2}+y\right)_{-1/2}^1-\frac{1}{6}(y^{3})^{1}_{-1/2}$

              =  $\frac{1}{4}\left\{\left(\frac{1}{2}+1\right)-\left(\frac{1}{8}-\frac{1}{2}\right)\right\}-\frac{1}{6}\left\{1+\frac{1}{8}\right\}$

       =   $\frac{1}{4}\left\{\frac{3}{2}+\frac{3}{8}\right\}-\frac{1}{6}\left\{\frac{9}{8}\right\}$

   =  $\frac{1}{4}\times \frac{15}{8}-\frac{3}{16}=\frac{9}{32}$

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