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How to find circle equation when given its center lies in a line?

Question:

The center of a circle lies in the line 2x + y + 6 = 0. The circle is tangent to both x-axis and y-axis. Find the equation of the circle.

Solution:

The line equation 2x + y + 6 = 0. When x = 0, y = -6. When y = 0, x = -3. So, the line passes these two points (0, -6), (-3, 0). Draw this line, find this line passes Quadrant II, Quadrant III and Quadrant IV.

If the center of the circle lies in Quadrant III, let C1 be center of the circle. C1 has coordinate (xo, yo). The circle is tangent to the x-axis at point A. Then C1A is perpendicular to the x-axis at point A and C1A = yo < 0. and r = |C1A|. The circle is tangent to the y-axis at point B. So, C1B perpendicular to the y-axis at B. C1B = xo < 0 and r = |C1B|. Because xo < 0, yo < 0, and |xo| = |yo| = r, we get xo - yo = 0 ... name this as equation1.

Because the center of the circle (xo, yo) lies in the line 2x + y + 6 = 0, so, (xo, yo) satisfy the line equation 2x + y + 6 = 0. Now we substitute (xo, yo) into the line equation, we get 2xo + yo + 6 = 0 ... name this as equation2.

xo - yo = 0 ... equation1
2xo + yo + 6 = 0 ... equation2
equation1 + equation2 to remove yo
3xo + 6 = 0
xo = -2
substitute the value of xo into equation1 to get yo
yo = xo = -2
r = |xo| = 2
So, in Quadrant II, the circle equation is: (x + 2)2 + (y + 2)2 = 4

If the center of the circle lies in Quadrant II, let C2 is the center of the circle. C2 has the coordinate (xo, yo). The circle is tangent to the x-axis at point E. So, C2E perpendicular to the x-axis at E and C2E = y0 > 0 and yo = r. The circle is tangent to the y-axis at point F. Then C2F = xo < 0 and |xo| = r. Because xo < 0, yo > 0 and |xo| = |yo| = r, so, xo + yo = 0 ... name this as equation1

Because the center of the circle (xo, yo) lies in the line 2x + y + 6 = 0, so, (xo, yo) satisfy the line equation 2x + y + 6 = 0. Now we substitute (xo, yo) into the line equation, we get 2xo + yo + 6 = 0 ... name this as equation2.

xo + yo = 0 ... equation1
2xo + yo + 6 = 0 ... equation2
equation1 - equation2 to remove yo
-xo - 6 = 0
xo = -6
substitute the value of xointo equation1 to get yo
yo = -xo = -(-6) = 6
r = |xo| = 6
So, in Quadrant II, the circle equation is: (x + 6)2 + (y - 6)2 = 36
how to find circle equation when given circle center in a line and circle tangent to both axis?

Look the figure above, the blue circle lies in Quadrant II and tangent to both x-axis and y-axis. The center of the blue circle has coordinate (-6, 6) that lies in the line 2x + y + 6 = 0. The orange circle lies in Quadrant III. The center of the orange circle has coordinate (-2, -2) that lies in the line 2x + y + 6 = 0. The orange circle tangent to both x-axis and y-axis. Watch the video for more details.