Show that the circles, x2+y2−4x−6y−12=0 and x2+y2+6x+18y+26=0 touch each other. Also find the point of contact and common tangent at this point of contact.
First circle - solve by completing the square:
x²+ y² - 4x - 6y - 12 = 0
(x² - 4x) + (y² - 6y) - 12 = 0
(x² - 4x + 4) + (y² - 6y + 9) - 25 = 0
(x-2)² + (y-3)² = 25
So this circle has its center at the point (2,3) and radius 5.
Do the same for the second circle:
x² + y² + 6x + 18y + 26 = 0
(x² + 6x) + (y² + 18y) + 26 = 0
(x² + 6x + 9) + (y² + 18y + 81) - 64 = 0
(x+3)² + (y+9)² = 64
So this circle has its center at the point (-3, -9) and radius 8.
How do we know they touch each other? The x coordinates differ by 5, the y coordinates differ by 12, and the sum of the two radii is 13, and 5/12/13 is a Pythagorean triplet. So the radii of the two circles form the hypotenuse of a right triangle, like this:
The point of tangency should be (+1/13, -21/13.) Since the slope of the line that connects the two radii is 12/5, the slope of the tangent line must be -5/12.
First circle - solve by completing the square:
x²+ y² - 4x - 6y - 12 = 0
(x² - 4x) + (y² - 6y) - 12 = 0
(x² - 4x + 4) + (y² - 6y + 9) - 25 = 0
(x-2)² + (y-3)² = 25
So this circle has its center at the point (2,3) and radius 5.
Do the same for the second circle:
x² + y² + 6x + 18y + 26 = 0
(x² + 6x) + (y² + 18y) + 26 = 0
(x² + 6x + 9) + (y² + 18y + 81) - 64 = 0
(x+3)² + (y+9)² = 64
So this circle has its center at the point (-3, -9) and radius 8.
How do we know they touch each other? The x coordinates differ by 5, the y coordinates differ by 12, and the sum of the two radii is 13, and 5/12/13 is a Pythagorean triplet. So the radii of the two circles form the hypotenuse of a right triangle, like this:
The point of tangency should be (+1/13, -21/13.) Since the slope of the line that connects the two radii is 12/5, the slope of the tangent line must be -5/12.
