Please help me on number 31 please I beg you
1 answer:
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Answer:
center = (3, 0)
radius = 2
Step-by-step explanation:
Standard equation of a circle:
(where (h, k) is the center of the circle and r is the radius)
Given equation:
Rewrite in standard form:
Therefore
- center = (3, 0)
- radius = 2
Answer:
c) 6x + y = -52 is required equation perpendicular to the given equation.
Step-by-step explanation:
If the equation is of the form : y = mx + C.
Here m = slope of the equation.
Two equations are said to be perpendicular if the product of their respective slopes is -1.
Here, equation 1 : -x + 6y = -12
or, 6y = -12 + x
or, y = (x/6) - 2
⇒Slope of line 1 = (1/6)
Now, for equation 2 to be perpendicular:
Check for each equation:
a. x + 6y = -67 ⇒ 6y = -67 - x
or, y = (-x/6) - (67/6) ⇒Slope of line 2 = (-1/6)
but
b. x - 6y = -52 ⇒ -6y = -52 - x
or, y = (x/6) + (52/6) ⇒Slope of line 2 = (1/6)
but
c. 6x + y = -52
or, y =y = -52 - 6x ⇒Slope of line 2 = (-6)
Hence, 6x + y = -52 is required equation 2.
d. 6x - y = 52 ⇒ -y = 52 - 6x
or, y = 6x - 52 ⇒Slope of line 2 = (6)
but
Hence, 6x + y = -52 is the only required equation .