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Tcecarenko [31]
2 years ago
7

This is so important please help

Mathematics
1 answer:
Shtirlitz [24]2 years ago
3 0

Given:

The center of the circle = (-1,0)

Point on the circle = (2,-4).

To find:

The equation of the circle.

Solution:

Distance between two points is

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance between the center (-1,0) and the point on the circle (2,-4) is the radius of the circle.

r=\sqrt{(2-(-1))^2+(-4-0)^2}

r=\sqrt{(2+1)^2+(-4)^2}

r=\sqrt{(3)^2+(-4)^2}

On further simplification, we get

r=\sqrt{9+16}

r=\sqrt{25}

r=5

Now, the equation of the circle is

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

Where, (h,k) is the center and r is the radius of the circle.

The radius of the circle is 5 and the center is at (-1,0), so the equation of the circle is

(x-(-1))^2+(y-0)^2=5^2

(x+1)^2+y^2=25

Therefore, the first option is correct, i.e., (x+1)^2+y^2=25.

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The question is ill formated, the complete question is

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<h3><u>Answer:</u></h3>

\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}

<h3><u>Step-by-step explanation:</u></h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

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\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2  \\\\\bf\implies (2y+3)^2 \leq 1  \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1  \\\\\bf\implies 4y^2+9+12y - 1 \leq 0  \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0  \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup   \\\\\bf\implies y^2y+2y+y+2 \leq 0  \\\\\bf\implies y(y+2)+1(y+2)\leq 0  \\\\\bf\implies ( y+2)(y+1)\leq 0  \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}

<h3><u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>
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