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kogti [31]
4 years ago
5

What is the equation of the circle with center (1,1) that passes through the point (-3,4)?

Mathematics
1 answer:
MakcuM [25]4 years ago
3 0

Answer:

(x - 1)2 + (y - 1)² = 5²

Step-by-step explanation:

Adapt the standard equation of a circle with center at (h, k) and radius r:

(x - 1)^2 + (y - 1)^2 = r^2

Here one point on the circle is (-3, 4), and so the radius is √[(-3)² + 4² ] = 5

Thus, the desired equation is:

(x - 1)^2 + (y - 1)^2 = 5^2      or     (x - 1)2 + (y - 1)² = 5²

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larisa [96]
<h3>Answer:   Slope = -3/7</h3>

===================================================

Apply the slope formula

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m = (6-3)/(-9-(-2))

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m = -3/7

This means each time you go down 3 units, you go to the right 7 units.

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3 years ago
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kati45 [8]

Answer:

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Step-by-step explanation:

A unit circle is a circle with a radius of 1 .Because the radius is 1, it is possible to directly measure the sine, cosine and tangent.

<em>using the unit circle where 90° is the limit as the hypotenuse approaches the vertical y-axis which is positive.</em>

Sine=opposite/hypotenuse

Sin=O/H

<u>Applying the limits</u>

Sine 90°=1/1= 1

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or

When the angle formed at the origin of the unit circle in the 1st quadrant is 0°, cos 0°=1 sin0°=0  and tan 0°=0

When we increase the angle until it is 90°, cos 90°=0, sin 90°=1 and tan 90°=undefined

 

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