Question

The centre of a circle is (x+1),(x-1).Find x if the circle passes through (2,2)and (8,-2)

Answer 1

Solution:

Given: The center of a circle has coordinates (x+1,x-1). The circle passes through (2,2) and (8,-2).

To find: value of x(x=?)

=>All the points that passes that through the circumference of a circle are equidistant from center. (radius are always equal).

Then,
According to distance formula,
 
               A(x,y)    =$\sqrt{(x_{2}-x_{1} ) ^{2} +(y_{2}- y_{1}) ^{2} } }$      
         $\sqrt{(x+1-2) ^{2}+(x-1-2) ^{2} }= \sqrt{(x+1-8) ^{2}+(x-1-(-2) ^{2} }$
        $\sqrt{(x-1)^2+(x-3)^2} = \sqrt{(x-7)^2+(x+1)^2}$
$\sqrt{x^2-2x+1+x^2-6x+9} = \sqrt{x^2-14x+49+x^2+2x+1}$$2x^2-8x+10=2x^2-12x+50$
-8x+12x+10-50=0
4x-40=0
4x=40
x=10        hope it helps