Question

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

Answer 1

#1

• Centre (-1,2)

Point on circle=(-3,4)

Radius :-

`\\ \rm\Rrightarrow √((-1+3)^2+(2-4)^2)`

`\\ \rm\Rrightarrow √(2^2+(-2)^2)`

`\\ \rm\Rrightarrow √(8)`

`\\ \rm\Rrightarrow 2√(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

Answer 2

Answers:

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

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

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

Step-by-step 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 = √((x_1 - x_2)^2 + (y_1 - y_2)^2)\\\\d = √((-1-(-3))^2 + (2-4)^2)\\\\d = √((-1+3)^2 + (2-4)^2)\\\\d = √((2)^2 + (-2)^2)\\\\d = √(4 + 4)\\\\d = √(8)\\\\`

This is the radius because it stretches from the center to a point on the circle, so
`r = √(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.

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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 = √((x_1 - x_2)^2 + (y_1 - y_2)^2)\\\\d = √((3-8)^2 + (13-25)^2)\\\\d = √((-5)^2 + (-12)^2)\\\\d = √(25 + 144)\\\\d = √(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.