What is an equation of a circle whose center is (1,4) and diameter is 10?

Mathematics

1 answer:

nexus9112 [7]3 years ago

6 0

Answer:

$x^{2} - 2x + y^{2} - 8y = 8$

Step-by-step explanation:

Standard form for a circle is:

$(x - h)^{2} + (y - k)^{2} = r^{2}$ (h, k) is the center r is the radius

$(x - 1)^{2} + (y - 4)^{2} = 5^{2} \\x^{2} - 2x + 1 +y^{2} - 8y + 16 = 25\\x^{2} - 2x + y^{2} - 8y = 8$ $x^{2} - 2x + y^{2} - 8y = 8$

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Match the inequality to the graph of its solution. a. n/4 ≤ -1 b. -10 ≥ -100 c. 5x ≥ 20

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a. n/4 ≤ -1

Multiply both sides by 4 => n ≤ - 4, which is all the real numbers less or equal than - 4.

That in the real number line is all the numbers to the left of - 4 (including -4)

The matching graph is the B.

b. -10n ≥ -100

Divide both sides by - 10 => n ≤ 10

That is all the real numbers less or equal than 10.

In the real number line it is all the numbers to the left of 10, including 10.

So, the matching graph is the A.

c. 5x ≥ 20

Divide both sides by 5 => x ≥ 4

That is all the real numbers greater or equal to 4.

In the real number line it is all the numbers to the right of 4, including 4.

The matching graph is C.

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