Find an equation of the circles passing through the points A 1 2 and B 3 4 and tangent to the line 3x plus y equals 3?

Answer 1

The given information is not enough to write the equation of a circle. So we need to work a little bit.

Plot A(1, 2) and B(3,4) in a rectangular coordinate system. Draw the tangent line, y = -3x + 3, which cuts the x-axis at (1, 0). By looking at the position of the points A, B, and at the path of the tangent line, think that the point where the tangent touches the circle could be (1, 0) or close to.

Plan: The center of the circle is at the intersection point of the perpendicular line to tangent at the point of tangency, and the perpendicular bisector of the segment AB (reason: a tangent to a circle is perpendicular to its radius; the perpendicular bisector of a chord passes through the center).

Solution:

Let's suppose that (1, 0) is the point of tangency.

Find the midpoint M of AB: M = ((1+3)/2, (2+4)/2) = (2, 3)

Find the slope of the line that passes though A and B: m = (4-2)/3-1) = 1

Find the equation of the perpendicular bisector of AB (whose slope must be -1 and passes through (2, 3)):

y = -1x + b; let (x, y) = (2, 3)

3 = -1*2 + b

b = 5

y = -x + 5

Find the equation of the perpendicular line to the tangent (whose slope is 1/3 since the perpendicular lines have negative reciprocal slopes) and passes through (1, 0)):

y = 1/3x + b; let (x, y) = (1, 0)

0 = (1/3)*1 + b

b = -1/3

y = (1/3)x - 1/3

Find the intersection point of y = -x + 5 and y = (1/3)x - 1/3

-x + 5 = (1/3)x - 1/3

(4/3)x = 16/3

x = 16/3 * 3/4

x = 4 implies y =1

So the point (4, 1) could be the center of the circle.

If we compute the distances between (4, 1) and (1, 0), (1, 2), (3,4), we see that they have the same length, √10, then we say the center of the circle is (4,1) and the radius has a length of √10.

Thus, the equation of the circle that passes through (1, 2) and (3, 4) is the equation of a circle with center (4, 1) and a radius of √10.

(x - 4)2 + (y - 1)2 = 10

Let's solve the problem without risking extra work (we clearly see that our guess for the point of tangency was right).

After we draw the given information on the rectangular system, we would like to draw a parallel line to the tangent line that passes through the point B(3, 4). For this we need the equation of the line to use its slope to draw it and to find the point of intersection with the perpendicular bisector of segment AB (since the center of the circle lies there; see above); (the slope is -3, since parallel line have the same slope).

y = -3x + b; ; let (x, y) = (3, 4)

4 = -3*3 + b

b = 13

y = -3x + 13

Let's find the intersection point of y = -x + 5 and y = -3x + 13

-x + 5 = -3x + 13

2x = 8

x = 4 implies y = 1, say O(4, 1)

Let's find the length of the congruent (O lies on the perpendicular bisector) segments OA and OB, maybe they are radii.

OA = OB = √((4-1)2+(1-2)2) = √10.

Since the point O lies in the same time on the perpendicular bisector of AB and on the parallel line to the tangent, let's find the equation of the perpendicular line to the tangent that passes through O (the slope is 1/3 since the perpendicular lines have negative reciprocal slopes):

y = (1/3)x + b; let (x, y) = (4, 1)

1 = (1/3)*4 + b

b = -1/3

y = (1/3)x - 1/3

Let's find the intersection point of y = -3x + 3 and y = (1/3)x - 1/3 (maybe this would be the point of tangency):

-3x + 3 = (1/3)x - 1/3

(4/3)x = 4/3

x = 1 implies y = 0; say T(1, 0)

Let's find the length of OT:

OT = √((4-1)2+(1-0)2) = √10.

Since OA, OB, and OT have the same length, they are radii of the circle with center O(4, 1). So our guess was right