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2 years ago

What is the center or a circle represented by the equation (x+9)2 + (y-6)2 = 10 to the power of 2

Mathematics

1 answer:

m_a_m_a [10]2 years ago

4 0

Answer:

(-9,6)

Step-by-step explanation:

The general equation of a circle is given as:

x²+y²+2gx+2fy+c = 0

The centre of the circle is (-g,-f)

Given the equation of the circle

(x+9)² + (y-6)² = 10²

First we need to expand and write it in the general form. On expansion:

x²+18x+81+y²-12y+36 = 100

Collecting like terms

x²+y²+18x-12y+81+36-100 = 0

x²+y²+18x-12y+17 = 0

Comparing to the general equation

2gx = 18x

g = 9

2fy = -12

f = -6

The center of the circle = (-g, -f)

= (-9,-(-6))

= (-9, 6)

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Answer:

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Step-by-step explanation:

Given

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2 years ago

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2 years ago

Find the perimeter of the pentagon MNPQR with vertices M(2, 4), N(5, 8), P(8, 4), Q(8, 1), and R(2, 1)

Gekata [30.6K]

Answer:

The pentagon MNPQR has a perimeter of 22 units.

Step-by-step explanation:

Geometrically speaking, the perimeter of the pentagon is the sum of the lengths of each side, that is:

$p = MN + NP + PQ + QR + RM$ (1)

$p = \sqrt{\overrightarrow{MN}\,\bullet \, \overrightarrow{MN}} + \sqrt{\overrightarrow{NP}\,\bullet \, \overrightarrow{NP}} + \sqrt{\overrightarrow{PQ}\,\bullet \, \overrightarrow{PQ}} + \sqrt{\overrightarrow{QR}\,\bullet \, \overrightarrow{QR}} + \sqrt{\overrightarrow{RM}\,\bullet \, \overrightarrow{RM}}$ (1b)

If we know that $M(x,y) = (2,4)$ , $N(x,y) = (5,8)$ , $P(x,y) = (8,4)$ , $Q(x,y) = (8,1)$ and $R(x,y) = (2,1)$ , then the perimeter of the pentagon MNPQR is:

$p =\sqrt{(5-2)^{2}+(8-4)^{2}} + \sqrt{(8-5)^{2}+(4-8)^{2}}+\sqrt{(8-8)^{2}+(1-4)^{2}}+\sqrt{(2-8)^{2}+(1-1)^{2}}+\sqrt{(2-2)^{2}+(4-1)^{2}}$ $p = \sqrt{3^{2}+4^{2}} + \sqrt{3^{2}+(-4)^{2}}+\sqrt{0^{2}+(-3)^{2}}+\sqrt{(-6)^{2}+0^{2}}+\sqrt{0^{2}+3^{2}}$

$p = 22$

The pentagon MNPQR has a perimeter of 22 units.

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2 years ago