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worty [1.4K]
3 years ago
13

Helppp i’ll give brainlest

Mathematics
1 answer:
frez [133]3 years ago
8 0

Answer:

\boxed{\textsf{ The center of the circle is $\bf \bigg( -\dfrac{17}{2},0\bigg)  $ .}}

Step-by-step explanation:

A equation of circle is given to us . From the equation we need to find out the centre of the circle .The equation is :-

\sf\implies x^2+y^2+17x+42=0

Now we know the standard equation of a circle as , ( x - h )² + (y - k)² = r² , where

  • (h , k) is the centre of the circle.
  • r is the radius of the circle .

Now , let's simplify out the equation and convert it into standard form .

\sf\implies x^2+y^2+17x+42=0

Step 1: <u>Subtract</u><u> </u><u>4</u><u>2</u><u> </u><u>from </u><u>both </u><u>sides </u><u>:</u><u>-</u><u> </u>

\sf\implies x^2+y^2+17x+42-42=0-42\\\\\sf\implies x^2+y^2+17x = -42

Step 2:<u>Group </u><u>out </u><u>the</u><u> </u><u>like </u><u>terms</u><u>:</u><u>-</u>

\sf\implies (x^2+17x)+y^2 = -42

Step 3: <u>Complete</u><u> the</u><u> </u><u>square </u><u>for </u><u>x²</u><u> </u><u>:</u><u>-</u><u> </u>

\sf\implies \bigg( x^2+\dfrac{2\times 17x}{2} + \bigg(\dfrac{17}{2}\bigg)^2\bigg)+y^2 = -42+\bigg(\dfrac{17}{2}\bigg)^2\\\\\sf\implies \bigg( x + \dfrac{17}{2}\bigg)^2 + y^2 =\dfrac{289}{4}-42

Step 4: <u>Convert </u><u>the </u><u>RHS </u><u>into </u><u>whole</u><u> </u><u>square</u><u>.</u>

\sf\implies \bigg( x + \dfrac{17}{2}\bigg)^2 + y^2 =\dfrac{289-168}{4}  \\\\\sf\implies\bigg( x + \dfrac{17}{2}\bigg)^2 + y^2 =\dfrac{121}{4}\\\\\sf\implies \bigg( x + \dfrac{17}{2}\bigg)^2 + y^2 =\bigg(\dfrac{11}{2}\bigg)^2

Step 6: <u>Convert </u><u>it </u><u>into </u><u>the </u><u>standard </u><u>form </u><u>.</u>

\sf\implies \bigg\{ x - \bigg( - \dfrac{17}{2}\bigg) \bigg\}^2+ ( y - 0 )^2 = \bigg(\dfrac{11}{2}\bigg)^2

Hence here we got our equation in the standard form . On comparing it to the standard form we get that ,

\sf\implies \pink{ Center = \bigg( -\dfrac{17}{2},0\bigg) }

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