Question

Find an equation of the circle, which crosses the x-axis at the points (2, 0) and (8,0) and touches the y-axis at the point (0, 4).​

Answer 1

Explanation:

please see the attached document for detailed explanation and the diagram showing the circle and the given two points.

As the line joining (2,0) and (8,0) is on the x axis, the center lies on the perpendicular bisector of the line segment joining the two points. So the center lies on x = (2+8)/2 or x = 5. As the y-axis is a tangent to the circle at (0,4), it means that the y-coordinate of the center is = 4.

So the center is (5,4).

Then the radius can be found as :

√[(5-2)²+(4-0)²] = 5

The equation of the circle is then :

(x-5)²+(y-4)² = 5²

x²-10x+y²-8x+16=0.

Answer 2

To find the equation of a circle, we need to determine the center and radius of the circle. Given that the circle crosses the x-axis at (2, 0) and (8, 0), we can determine that the center of the circle lies on the line y = 0 (since the y-coordinate of the center is 0).

Since the circle also touches the y-axis at (0, 4), we know that the distance between the center of the circle and the y-axis is equal to the radius of the circle. Therefore, the radius is 4 units.

Since the y-coordinate of the center is 0, and the radius is 4, we can determine that the center of the circle is at (0, 0).

Now, we can use the standard equation of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.

Substituting the values we have, we get (x - 0)^2 + (y - 0)^2 = 4^2.

Simplifying further, we have x^2 + y^2 = 16.

Therefore, the equation of the circle is x^2 + y^2 = 16.