Answer:
The circles touch externally.
Step-by-step explanation:
The general equation of a circle is given by;
x² + y² + 2gx + 2fy + c = 0 -------------(i)
Where;
g, f and c are all constants
and;
(-g, -f) gives the centre of the circle
√(g² + f² - c) gives the radius of the circle.
Now compare the given circles with the general equation to find the centres and radii of the two circles.
(i) For circle 1
x²+y²=100
This can be re-written as
x²+y² + 2(0)x + 2(0)y - 100 = 0 --------------(ii)
Comparing equation (ii) with equation (i), we see that;
g = 0
f = 0
c = -100
Therefore,
(a) the centre (-g, -f) of the circle is (0,0)
(b) the radius √(g² + f² - c) of the circle is
=> √(0² + 0² - (-100))
=> √(100)
=> 10
(i) For circle 2
x²+y²-24x -10y +160=0
This can be re-written as
x²+y² + 2(-12)x + 2(-5)y + 160 = 0 --------------(iii)
Comparing equation (iii) with equation (i), we see that;
g = -12
f = -5
c = 160
Therefore,
(a) the centre (-g, -f) of the circle is (12,5)
(b) the radius √(g² + f² - c) of the circle is
=> √((-12)² + (-5)² - (160))
=> √(144 + 25 - 160)
=> √9
=> 3
(iii) Two circles touch externally if the distance between their centres is equal to the sum of their radii.
(a) Let's calculate the distance (d) between the centres of the circles as follows;
d = √(x₂ - x₁)² + (y₂ - y₁)²
Where;
x₂ and x₁ are the x-coordinates of the centres of the first and second circles respectively i.e x₂ = 12 and x₁ = 0
y₂ and y₁ are the y-coordinates of the centres of the first and second circles respectively i.e y₂ = 5 and y₁ = 0
=> d = √(12 - 0)² + (5 - 0)²
=> d = √(12)² + (5)²
=> d = √169
=> d = 13
(b) Let's calculate the sum (s) of the radii of the two circles.
s = 10 + 3
s = 13
Now, since the distance between their centres is equal to the sum of their radii, the two circles does touch externally.
