Question

find the equation of the circle with centre on the X axis and passing through the origin and having radius 4​

Answer 1

So centre is on x axis that means x coordinate is equal to radius
The formula for circle is (x-a)^2+(y-b)^2=r^2
We have a =4 b=0 r=4
(x-4)^2+(y-0)^2=(4)^2
x^2 +16-8x +y^2 =16
x^2+y^2-8x=0 is the equation of circle

Answer 2

The equation of circle is $x^2-8x+y^2=0$.

Step-by-step explanation:

Consider the provided information.

The equation of the circle with centre on the X axis and passing through the origin and having radius 4

Since, circle passes through the origin and centre lies on x-axis then the abscissa will be equal to the radius of the circle and the y co-ordinate of the centre will be zero.

The equation of the circle is $(x - h)^2 + (y - k)^2 = a^2$

Where a is the radius with centre at (h, k).

Since, the centre is on x axis the value of k is zero.

The required equation of the circle passes through the origin and having radius 4​ is:

$(x-4)^2+(y-0)^2=4^2$

$x^2-8x+y^2=0$

Hence, the equation of the circle is $x^2-8x+y^2=0$.

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