Question

Find the equation of the circle which
passes through (1, 1) and cuts orthogonally
each of the circles x² + y2 – 8x - 2y + 16 = 0
and x² + y2 - 4x - 4y-1=0.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$x^{2} +y^{2} -\frac{14}{3}x+\frac{23}{3} y-5=0$ of the circle which passes through $(1,1)$ and cuts orthogonally each of the circles $x^{2} +y^{2} -8x-2y+16=0$ and $x^{2} +y^{2} -4x-4y-1=0$

step 1: To find the $g,f,c$ values by using the circle equation

Let we consider the circle  $S=x^{2} +y^{2} +2gx+2fy+c=0$  passes through $(1,1)$

let apply  $(x,y)=(1,1)$  in the above equation

$x+y+2gx+2fy+c=0$  this is the first equation.

$1+1+2g+2f+c=0\\2+2g+2f+c=0\\c=-2g-2f-2$

the equation 1 orthogonally to  $x^{2} +y^{2} -8x-2y+16=0$

$g_{1}=g\\f_{1} =f\\c_{1}=-2g-2f-2$

$g_{2}=-4\\f_{2} =-1\\c_{2}=16$

$c_{2}=16$ get in the equation of $x^{2} +y^{2} -8x-2y+16=0$

let apply the known values in the equation $2g_{1}g_{2} +2f_{1}f_{2}=c_{1}+c_{2}$

$2g(-4)+2f(-1)=-2g-2f-2+16\\-8g-2f=-2g-2f+14\\-8g-2f+2g+2f=14\\-8g+2g=14$

$-6g=14\\g=-\frac{14}{6}\\g=-\frac{7}{3}$

this  $g=-\frac{7}{3}$  is 2nd equation.

then, the equation 1, orthogonally to  $x^{2} +y^{2} -4x-4y-1=0$

$g_{1}=g\\f_{1} =f\\c_{1}=-2g-2f-2$

$g_{2}=-2\\f_{2} =-2\\c_{2}=-1$

$c_{2}=-1$ get in the equation of  $x^{2} +y^{2} -4x-4y-1=0$

let apply the known values in the equation  $2g_{1}g_{2} +2f_{1}f_{2}=c_{1}+c_{2}$

$2g(-2)+2f(-2)=-2g-2f-2-1\\-4g-4f=-2g-2f-3\\-4g-4f+2g+2f=-3\\-2g-2f=-3\\2g+2f=3$

this is equation 3

from the equation 2,  $g=-\frac{7}{3}$  apply in the equation 3

$2g+2f=-3\\2(-\frac{7}{3} )+2f=3\\-\frac{14}{3} -3=-2f\\\frac{-14-3}{3} =-2f$

$\frac{-23}{3} =-2f\\\frac{-23}{3} =-2f\\\frac{23}{6}=f$

we get the values  $f=\frac{23}{6},g=\frac{-7}{3}$  apply the equation  $c =-2g-2f-2$

$c=-2(\frac{-7}{3}) -2(\frac{23}{6} )-2\\c=\frac{14}{3} -\frac{46}{6}-2\\c=\frac{28-46-12}{6} \\c=\frac{-30}{6} \\c=-5$

step 2: To find which equation passes through $(1,1)$ and cuts orthogonally each of the circles $x^{2} +y^{2} -8x-2y+16=0$  and $x^{2} +y^{2} -4x-4y-1=0$

finally we get the values $g,f,c$ and apply the circle equation $S=x^{2} +y^{2} +2gx+2fy+c=0$

$x^{2} +y^{2} +2gx+2fy+c=0\\x^{2} +y^{2}+2(-\frac{7}{3} )x+2(\frac{23}{6} )y+(-5)=0\\x^{2} +y^{2} -\frac{14}{3}x+\frac{23}{3} y-5=0$

the equation  $x^{2} +y^{2} -\frac{14}{3}x+\frac{23}{3} y-5=0$  of the circle which passes through  $(1,1)$ and cuts orthogonally each of the circles  $x^{2} +y^{2} -8x-2y+16=0$  and   $x^{2} +y^{2} -4x-4y-1=0$