Find the equation of the circle which touches the line y is equal to 2 and passes through origin and the point where the curve y square minus 2 times x plus eight is equal to 0 meets x axis?

Answer 1

The curve y² -2x + 8 = 0 meets the x-axis when y = 0; thus:

0² - 2x + 8 = 0

→ 2x = 8

→ x = 4.

The circle thus passes through the two points (0, 0) and (4, 0); it also has a tangent at y = 2; thus:

The chord of the circle between the two points (0, 0) and (4, 0) is parallel to the tangent (at y = 2), thus the x-coordinate of the point when the tangent at y = 2 touches the circle is the midpoint of the chord, namely (4 + 0) / 2 = 2;

Thus three points on the circle are (0, 0), (4,0) and (2, 2).

The general equation of a circle with centre (x0, y0) and radius r is (x - x0)² + (y - y0)² = r²; thus:

As all three points must satisfy this equation:

1) point (0, 0):

(0 - x0)² + (y - y0)² = r²

→ x0² + y0² = r²

2) point (4, 0):

(4 - x0)² + (0 - y0)² = r²

→ 16 - 8x0 + x0² + y0² = r²

But from (1) x0² + y0² = r²

→ 16 - 8x0 + r² = r²

→ 16 - 8x0 = 0

→ 8x0 = 16

→ x0 = 2

2) point (2, 2):

(2 - x0)² + (2 - y0)² = r²

→ 4 - 4x0 + x0² + 4 - 4y0 + y0² = r²

→ 8 - 4x0 - 4y0 + x0² + y0² = r²

But from (1) x0² + y0² = r², and from (2) x0 = 2

→ 8 - 4 × 2 - 4y0 + r² = r²

→ 0 - 4y0 = 0

← y0 = 0

Thus the centre of the circle is at (2, 0); and using (1) the radius (squared) can be found:

x0² + y0² = r²

→ 2² + 0² = r²

→ r² = 4

Thus the equation of the circle is:

(x - 2)² + (y - 0)² = 2²

→ x² -4x + 4 + y² = 4

→ x² + y² - 4x = 0

The above can be shortened by noting that as the tangent at y = 2 is parallel to the chord between the other two points (as both are given to have a y-coordinate of 0), not only is its x-coordinate the mid point of the two points on the chord on the x-axis, that is also the x-coordinate of the centre of the circle as the chord is horizontal, the line from the point where the tangent touches the circle to the centre of the circle is going to be vertical, ie a line of the form x = something.