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Bogdan [553]
3 years ago
13

If the endpoints of the diameter of a circle are (8, −6) and (4, −2), what is the standard form equation of the circle?

Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0
We do the midpoint formula to find the center of the circle to get the left side of the equation.

Midpoint = [(X₁ + X₂) / 2 , (Y₁ + Y₂) / 2] 

Now plug in:

[(8 + 4) / 2 , (- 6 - 2) / 2]
(12 / 2 , - 8 / 2)
(6, - 4)

The center of the circle is (6, -4)

Now we plug it into the equation of a circle:

(x - h)² + (y - k)² = r² where (h, k) is the center of the circle and r is the radius. 

(x - 6)² + (y + 4)² = r² is the left side of the equation. This will eliminate options A and C

Now we do the distance formula using the center and an endpoint to get the radius. The formula for distance is:

√((X₂ - X₁)² + (Y₂ - Y₁)²)

We plug in using either of the endpoints. 

√((4 - 6)² + (- 2 - (- 4))²)
√((-2)² + 2²)
√(4 + 4)
√8

√8 is your radius

(√8)² = 8

Your correct answer is (x - 6)² + (y + 4)² = 8, Option D
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If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha
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6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-\frac{2}{3}\\\\2x-1=0\to x=\frac{1}{2}\\\\\alpha=-\frac{2}{3};\ \beta=\frac{1}{2}\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-\frac{2}{3}}{\frac{1}{2}}+\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{2}{3}\cdot\frac{2}{1}-\frac{1}{2}\cdot\frac{3}{2}=-\frac{4}{3}-\frac{3}{4}\\\\=-\frac{16}{12}-\frac{9}{12}=-\frac{25}{12}=-2\frac{1}{12}


use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2

6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{1^2}{6\cdot(-2)}-2=\frac{1}{-12}-2=-2\frac{1}{12}
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3 years ago
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