.hello :
an equation of the circle <span>Center at the w(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : r = 1
</span><span>The points (-18,15) and (-20,15) lie on a circle with a radius of 1:
</span>(-18-a)²+(15-b)² = 1 ....(1)
(-20-a)² +(15-b)² = 1 ....(2)
solve this system :
(1) -(2) : (-18-a)² - (-20-a)² =0
(-18-a)² =(-20-a)² =0
( -18-a = -20-a) or (-18-a = - (-20-a))
1 ) ( -18-a = -20-a) no solution confused : -18=-20
2 ) -18-a =20+a
-2a =38
a = -19
subst in (1) :(-18+19)²+(15-b)² =1
(15-b)² = 0.... 15-b = 0 .... b = 15
the center is :w(-19,15)
Answer:
y=3x+3
Step-by-step explanation:
Answer:
Step-by-step explanation:
Domain:all real values
Range:[5,∞)
What do u need help with ?, trigonometry?
it's plural so axes*
and in pilar coordinates, the axes are circular so they never intersect but the centre point is still called origin.
