Math, asked by pallapurejyothi, 7 hours ago

Find the centre and radius of the circle
represented by :
(a^2-40)x^2+ (a-b -50)xy+(b^2-5)y^2+8x+8y+8=0 ​

Answers

Answered by jjinal231
1

Step-by-step explanation:

(x−h)

2

+(y−k)

2

=r

2

left parenthesis, x, minus, start color #11accd, h, end color #11accd, right parenthesis, squared, plus, left parenthesis, y, minus, start color #ca337c, k, end color #ca337c, right parenthesis, squared, equals, start color #e07d10, r, end color #e07d10, squared

This is the general standard equation for the circle centered at (\blueD h, \maroonD k)(h,k)left parenthesis, start color #11accd, h, end color #11accd, comma, start color #ca337c, k, end color #ca337c, right parenthesis with radius \goldD rrstart color #e07d10, r, end color #e07d10.

Circles can also be given in expanded form, which is simply the result of expanding the binomial squares in the standard form and combining like terms.

For example, the equation of the circle centered at (\blueD 1,\maroonD 2)(1,2)left parenthesis, start color #11accd, 1, end color #11accd, comma, start color #ca337c, 2, end color #ca337c, right parenthesis with radius \goldD 33start color #e07d10, 3, end color #e07d10 is (x-\blueD 1)^2+(y-\maroonD 2)^2=\goldD 3^2(x−1)

2

+(y−2)

2

=3

2

left parenthesis, x, minus, start color #11accd, 1, end color #11accd, right parenthesis, squared, plus, left parenthesis, y, minus, start color #ca337c, 2, end color #ca337c, right parenthesis, squared, equals, start color #e07d10, 3, end color #e07d10, squared. This is its expanded equation:

\begin{aligned} (x-\blueD 1)^2+(y-\maroonD 2)^2&=\goldD 3^2 \\\\ (x^2-2x+1)+(y^2-4y+4)&=9 \\\\ x^2+y^2-2x-4y-4&=0 \end{aligned}

(x−1)

2

+(y−2)

2

(x

2

−2x+1)+(y

2

−4y+4)

x

2

+y

2

−2x−4y−4

=3

2

=9

=0

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