All circles have the general equation
(x - h)² + (y - k) = r² with center (h,k) and radius r
Therefore every circle with center (1,4) has the equation
(x - 1)² + (y - 4) = r²
Three such circles with center (1,4) are drawn below:
Notice that the green circle intersects the line y=x in two points,
and the blue circle does not intersect the line y=x at all.
We are looking for the black circle, which intersects the line y=x
in exactly one point.
If we solve the system
by substituting x for y in the circle's equation:
(x - 1)² + (x - 4) = r²
x² - 2x + 1 + x² - 8x + 16 = r²
2x² - 10x + 17 = r²
2x² - 10x + 17 - r² = 0
2x² - 10x + (17-r²) = 0
We want this to have exactly one solution, so that the
line will be tangent to the circle. This will
be the case when and only when the discriminant b²-4ac
equals to 0.
Discriminant = b²-4ac = (-10)²-4(2)(17-r²) = 0
100 - 8(17-r²) = 0
100 - 136 + 8r² = 0
-36 + 8r² = 0
8r² = 36
r² =
r² =
That's all we need to form the equation of the circle:
(x - 1)² + (y - 4)² = r²
(x - 1)² + (y - 4)² =
x² - 2x + 1 + y² - 8y + 16 =
2x² - 4x + 2 + 2y² - 16y + 32 = 9
2x² + 2y² - 4x - 16y + 25 = 0
You aren't asked for the radius r but if you were asked for it
it would be:
r =
r =
r =
r =
Edwin
