show that the circles x^2 + y^2 - 10x - 8y + 18 = 0 and x^2 + y^2 - 8x - 4y + 14x = 0 do not intersect.
Apologies on behalf of my friend
I suspect still a typo, since there are two x terms.
Sorry, friend.
Sounds like a repeat of this without correcting the error.
http://www.jiskha.com/display.cgi?id=1493740391
No problem! To show that two circles do not intersect, we can compare the distances between their centers and the radii of the circles.
First, let's rewrite the equations of the circles in the standard form:
Circle 1: x^2 + y^2 - 10x - 8y + 18 = 0
Completing the square, we get: (x^2 - 10x + 25) + (y^2 - 8y + 16) = -18 + 25 + 16
Simplifying, we have: (x - 5)^2 + (y - 4)^2 = 23
Circle 2: x^2 + y^2 - 8x - 4y + 14 = 0
Completing the square, we get: (x^2 - 8x + 16) + (y^2 - 4y + 4) = -14 + 16 + 4
Simplifying, we have: (x - 4)^2 + (y - 2)^2 = 6
Now, we can compare the centers and radii of the two circles:
The center of Circle 1 is (5, 4), and the radius is √23.
The center of Circle 2 is (4, 2), and the radius is √6.
To show that the circles do not intersect, we need to prove that the distance between the centers is greater than the sum of the radii:
Distance between centers: √[(5 - 4)^2 + (4 - 2)^2] = √(1 + 4) = √5
Sum of the radii: √23 + √6
Since √5 is greater than √23 + √6, the circles do not intersect.
Therefore, we have shown that the circles x^2 + y^2 - 10x - 8y + 18 = 0 and x^2 + y^2 - 8x - 4y + 14x = 0 do not intersect.