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3 years ago

Suppose (12, 8) and (4, 6) are the endpoints of a diameter of a circle. Which is an equation of the circle?

Mathematics

1 answer:

In-s [12.5K]3 years ago

3 0

Answer:

$(x-8)^2 + (y-7)^2 - 17 = 0$

Step-by-step explanation:

The formula for the equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$

where (h,k) is the center and r is the radius of the circle.

To first solve solve this, use the endpoints of the diameter to solve for the center of the circle using the midpoint formula, which is known as:

$\frac{(x_1+x_2)}{2},\frac{(y_1+y_2)\ }{2}$

The points given are (12,8) and (4,6), so plug in the coordinates into the formula.

Center = (1/2)(12+4),(1/2)(8+6)

= (8,7)

Now, to find the radius, we can use the distance formula and use the center (8,7) and one of the endpoints, (12,8). The distance formula is known as:

$\sqrt{(x_{1} -x_{2})^2 +(y_{1} -y_{2})^2 }$

Radius = √(x₂- x₁)² + (y₂-y₁)²

= √(12- 8)² + (8-7)²

= √(4² + 1²) = √(16+ 1) = √17

= r² = 17

Now that the radius and the center are known, the equation of the circle can be written as:

$(x-8)^2 + (y-7)^2 - 17 = 0$

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Step-by-step explanation:

Given:

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Formula:

$cos(a-b)=cos(a)\ cos(b)+sin(a)\ sin(b)$

Plugging the values.

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So,

⇒ $cos(45-60)=(\frac{1}{\sqrt{2} } \times \frac{1}{2} ) + (\frac{1}{\sqrt{2} } \times \frac{\sqrt{3} }{2})$

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⇒ $cos(45-60)=(\frac{1+\sqrt{3} }{2\sqrt{2} } )\times \frac{2\sqrt{2} }{2\sqrt{2} }$ ...rationalizing

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⇒ $cos(45-60)=\frac{(\sqrt{2}+\sqrt{6})}{4}$

To find the exact values we have to put the values of sq-rt .

As, $\sqrt{2}=1.41$ and $\sqrt{6} =2.44$

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⇒ $cos(45-60)=0.96$

So the exact value of Cos(45° - 60°) is 0.96 using difference of two angles.

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