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blagie [28]
3 years ago
14

NO LINKS!!! Please help me with these problems​

Mathematics
2 answers:
nadezda [96]3 years ago
8 0
<h3>Answers:</h3>

7) Center= (-1,2) Radius= \boldsymbol{\sqrt{8}} Equation: (x+1)^2+(y-2)^2 = 8

8) Center= (3,13) Radius= 13 Equation: (x-3)^2+(y-13)^2 = 169

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

Explanation:

Problem 7

Let's find the distance from (-1,2) to (-3,4)

(x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\

This is the radius because it stretches from the center to a point on the circle, so r = \sqrt{8}

Squaring both sides will get us r^2 = 8

One useful template for a circle is the equation (x-h)^2+(y-k)^2 = r^2\\\\

(h,k) is the center

r is the radius

Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.

(x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\

You can confirm this by using a tool like Desmos. See below.

------------------------------------------------------------------------

Problem 8

The endpoints of the diameter are (-2,1) and (8,25)

The center is the midpoint of these endpoints.

The midpoint of the x coordinates is (-2+8)/2 = 3

The midpoint of the y coordinates is (1+25)/2 = 13

The center is (h,k) = (3,13)

Now find the distance from the center to one of the points on the circle, let's say to (8,25)

(x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\

The radius is exactly 13 units.

So,

(x-h)^2+(y-k)^2 = r^2\\\\(x-3)^2+(y-13)^2 = 13^2\\\\(x-3)^2+(y-13)^2 = 169\\\\

is the equation of this particular circle.

Visual confirmation is shown below.

Mnenie [13.5K]3 years ago
5 0

#1

  • Centre (-1,2)

Point on circle=(-3,4)

Radius :-

\\ \rm\Rrightarrow \sqrt{(-1+3)^2+(2-4)^2}

\\ \rm\Rrightarrow \sqrt{2^2+(-2)^2}

\\ \rm\Rrightarrow \sqrt{8}

\\ \rm\Rrightarrow 2\sqrt{2}

Now equation

\\ \rm\Rrightarrow (x-h)^2+(y-k)^2=r^2

\\ \rm\Rrightarrow (x+1)^2+(y-2)^2=8

#2

Endpoints of diameter

  • (-2,1)
  • (8,25)

. Midpoint is the centre

  • (-2+8/2,1+25/2)
  • (6/2,26/2)
  • (3,13)

Diameter

  • √(-2-8)²+(1-25)²
  • √(-10)²+(-24)²
  • √26²
  • 26

Radius=26/2=13

Equation

  • (x-3)²+(y-13)²=169
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