Question

find the equation of circle which passes through the point (7, 0) and (0 ,root7) and has it centre on x-axis​

Answer 1

The equation of the circle which passes through the point (7, 0) and (0, root7) and has its center on the x-axis​ is $(x-3)^{2} +(y-0)^{2} =16$.

Step-by-step explanation:

We have to find the equation of the circle which passes through the point (7, 0) and (0, root7) and has its center on the x-axis​.

Let the radius of the circle be 'r' and the center of the circle will be at the point (x, 0) because at the x-axis, the y-coordinate is 0.

As we know that the distance between the points through which the circle passes and the center of the circle is the radius of the circle.

So, $(x-x_1)^{2} +(y-y_1)^{2} = r^{2}$

     $(x-7)^{2} +(0-0)^{2} = r^{2}$     {because (7, 0) is the first point}

     $(x-7)^{2} = r^{2}$ ------------------- [equation 1]

Similarly, $(x-x_2)^{2} +(y-y_2)^{2} = r^{2}$

     $(x-0)^{2} +(0-\sqrt{7} )^{2} = r^{2}$     {because (7, 0) is the first point}

     $x^{2} +7 = r^{2}$ ------------------- [equation 2]

From equation 1 and 2 we get;

       $(x-7)^{2} = x^{2} +7$

       $x^{2} +49-14x = x^{2} +7$

        $14x= 49-7$

         $x=\frac{42}{14}$ = 3

Hence, the center of the circle is at point (3, 0).

Now, using equation 2, the value of radius of the circle is;

               $x^{2} +7 = r^{2}$

               $r^{2} =3^{2} +7 = 16$

So, the equation of the circle is given by;

$(x-3)^{2} +(y-0)^{2} =16$.