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).

Answers: 2

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

answered: Guest

175

step-by-step explanation:

2 inches x 10 miles = 200 miles - 3 inches = 175

answered: Guest

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation: