Question

if normal at p(1,-2) to the circle intersects it again at Q and equation of tangent at Q to circle is 4y=3x+14 then equation of circle is​

Answer 1

Answer:

don't know maybe my brother can get the answer

Answer 2

Answer:

The circle equation is $X^{2} +Y^{2} +X-6=0$

Step-by-step explanation:

TO FIND:  The equation of circle.

GIVEN:

       The point of normal the circle intersects is p(1,-2)

          The equation of tangent at Q to circle is 4Y=3X+14---(1)

Slope of tangent at point Q is $Y=\frac{3}{4} X+\frac{14}{4}$

                                          Slope = $\frac{3}{4}$

                    The slope of PQ will be  (M)= $\frac{-1}{\frac{3}{4} }$

                                                  M=$-\frac{4}{3}$

           The equation of PQ is $(Y-Y_{1} )= M*(X-X_{1} )$

                                                 

                                                  $(Y-(-2) )=-\frac{4}{3} (X-1 )$

                                                 $3(Y+2)=-4(X-1)$

                                                  $3Y+6=-4X+4$

                                                   $4X+3Y+2=0$-----(2)

                        For the equation Q  equations from  (1) and (2) then we get

                                   

                                     $4Y-3X-14=0\\4X+3Y+2=0$

                                $Y=\frac{3}{4} X+\frac{14}{4}$

                         $4X+3(\frac{3}{4} X+\frac{14}{4})+2=0$

                          $16X+3(3X+14)+8=0$

                           $16X+9X+42+8=0\\25X+50=0\\25X=-50\\X=-2$

    SO $Y=\frac{3}{4} X+\frac{14}{4}$ substituting X=-2 In this equation

          $Y=\frac{3}{4} (-2)+\frac{14}{4}$

          $Y=2$

The points of Q is (-2,2)

      The mid point of the points PQ of the circle is

              $c=(h,k)$

             $C=(\frac{1-2}{2} ,\frac{-2+2}{2} )$

              $C=(\frac{-1}{2} ,0)$  ,h=-1/2,k=0.

Now the circle equation is $(X-h)^{2} +(Y-k)^{2} =r^{2}$-----(3)

                  To find the radius of the circle,The normal of the circle passes through the center.

             2r=distance of p(1,-2)  from 3X-4Y+14=0 IS

                  $2r=(\frac{3*1-4*(-2)+14}{\sqrt{3^{2}+ 4^{2} } } )$

                    $2r=(\frac{3+8+14}{5} )$

                    $2r=\frac{25}{5} \\2r=5\\r=\frac{5}{2}$

substituting r value in equation (3)

     $(X-\frac{1}{2} )+(Y-0)=(\frac{5}{2})^{2} }$

     $X^{2} +\frac{1}{2} +X+Y^{2} =\frac{25}{4} \\X^{2}+ Y^{2}+X-6=0$

The circle equation is $X^{2} +Y^{2} +X-6=0$

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