What is the center of the circle with the equation: X2 + y2 +18x + 14y + 105 = 0

Mathematics

1 answer:

professor190 [17]3 years ago

8 0

Answer:

(9, -7)

Step-by-step explanation:

$x^2+y^2+18x+14y+105=0\\$

Rearrange like x and y terms next to each other, then complete the square on both.

$(x-9)^2-361 + (y+7)^2 - 49 + 105 = 0$

The negative of the # numbers instead the brackets give you the centre of the circle.

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<h2>Angles</h2>

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Explanation:

The relationship between ∠1 and ∠2 are supplementary angles, so when you add up their measurements, it will become 180°. Simply subtract 180 and 140 to get the measure of ∠2. As well as ∠3, they're linear pairs. And they are also supplementary. To determine the measure of ∠6 and ∠7, notice the relationship between ∠2 and ∠6. As you noticed, it is corresponding angles. So they have the same measurement. If ∠2 = 40°, then ∠6 = 40°. As well as ∠7, because the relationship between ∠6 and ∠7 are vertical pairs. So the angle measurement of ∠7 is also 40°.

Meanwhile, the relationship between ∠1 and ∠4 are vertical pairs. It means they also have the same measurement. So ∠4 = 140°. The relationship between ∠1 and ∠5 are corresponding angles, so they also have the same measurement. If ∠1 = 140°, then ∠5 = 140°. The relationship between ∠1 and ∠8 are alternate exterior angles, and they also have the same measurement. If ∠1 = 140°, then ∠8 = 140°.

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