Mathematics, 27.10.2020 17:50 mollymay125

WILL GIVE BRAINLIEST ! (WORTH THIRTY POINTS) The equation of a circle is (x−2)^2+(y+6)^2=100. RECALL! a) Find the equation of a circle that is externally tangent to the given circle and

has a center at (18, −6).

b) Find the equation of a circle that is externally tangent to the given circle and has a center at (14, 3).

c) Find the equation of a circle that is internally tangent to the given circle and has a center at (2, −10).

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Answer from: ktprada

a) The equation of a circle that is externally tangent to the given circle is $(x-18)^{2}+(y+6)^{2} = 6^{2}$ .

b) The equation of a circle that is externally tangent to the given circle is $(x-14)^{2}+(y-3)^{2} = 5^{2}$ .

c) The equation of a circle that is internally tangent to the given circle is $(x-2)^{2}+(y+10)^{2} = 6^{2}$ .

Step-by-step explanation:

a) From Analytical Geometry we must remember that if two circles are tangent to each other, then centers are both collinear and radii are segments of such line segment. Distance between centers (), dimensionless, are found by the following Pythagorean identity:

$d = \sqrt{(h_{2}-h_{1})^{2}+(k_{2}-k_{1})^{2}}$ (Eq. 1)

Where:

$h_{1}$ , $k_{1}$ - Coordinates of the center of the first circle, dimensionless.

$h_{2}$ , $k_{2}$ - Coordinates of the center of the second circle, dimensionless.

In addition, circles are represented by the following equation in standard form:

$(x-h)^{2}+(y-k)^{2} = r^{2}$ (Eq. 2)

Where is the radius of the circle.

From statement we know the following information of the two circles:

First circle

$(x-2)^{2}+(y+6)^{2} = 10^{2}$

Second circle

$(x-18)^{2}+(y+6)^{2} = r_{2}^{2}$

If we know that $C_{1} (x, y) = (2, -6)$ and $C_{2} (x,y) = (18,-6)$ , then the distance between centers is:

$d = \sqrt{(18-2)^{2}+[(-6)-(-6)]^{2}}$

d = 16

Then, the radius of the second circle is:

$r_{2} = d-r_{1}$ (Eq. 3)

( d = 16 , $r_{1} = 10$ )

$r_{2} = 16-10$

$r_{2} = 6$

Therefore, the equation of a circle that is externally tangent to the given circle is $(x-18)^{2}+(y+6)^{2} = 6^{2}$ .

b) From statement we know the following information of the two circles:

First circle

$(x-2)^{2}+(y+6)^{2} = 10^{2}$

Second circle

$(x-14)^{2}+(y-3)^{2} = r_{2}^{2}$

If we know that $C_{1} (x, y) = (2, -6)$ and $C_{2} (x,y) = (14,3)$ , then the distance between centers is:

$d = \sqrt{(14-2)^{2}+[3-(-6)]^{2}}$

d = 15

Then, the radius of the second circle is:

( d = 15 , $r_{1} = 10$ )

$r_{2} = 15-10$

$r_{2} = 5$

Therefore, the equation of a circle that is externally tangent to the given circle is $(x-14)^{2}+(y-3)^{2} = 5^{2}$ .

c) From statement we know the following information of the two circles:

First circle

$(x-2)^{2}+(y+6)^{2} = 10^{2}$

Second circle

$(x-2)^{2}+(y+10)^{2} = r_{2}^{2}$

If we know that $C_{1} (x, y) = (2, -6)$ and $C_{2} (x,y) = (2,-10)$ , then the distance between centers is:

$d = \sqrt{(2-2)^{2}+[(-10)-(-6)]^{2}}$

d = 4

Then, the radius of the second circle is:

$r_{2} = r_{1}-d$ (Eq. 4)

( d = 4 , $r_{1} = 10$ )

$r_{2} = 10-4$

$r_{2} = 6$

Therefore, the equation of a circle that is internally tangent to the given circle is $(x-2)^{2}+(y+10)^{2} = 6^{2}$ .

Answer from: Quest

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Answer from: Quest

sample answer: you can determine whether each element of the domain is paired with exactly one element of the range. for example, if given a graph, you could use the vertical line test; if a vertical line intersects the graph more than once, then the relation that the graph represents is not a function.

Is the relation shown a function? use the vertical line test, explain

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