Question

equation of the circle having center (a,a) and point (h,k)​

Answer 1

Answer:

x²+y²-2a(x+y)+2a(h+k)-(h²+k²)=0

Step-by-step explanation:

Given --->

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Coordinates of centre of circle =(a,a)

Coordinates of point on circle=(h,k)

To find---> Equation of circle

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Solution--->

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Let centre of circle be O and point on circle be P

So Coordinate of O =(a,a)

Coordinate of P=(h,k)

Distance between centre and any point on circle is equal to radius

So radius =OP

=√{(a-h)²+(a-k)²} ,here square root is on whole expression

Equation of circle having centre at (α,β)and radiua r is

(x-α)²+(y-β)²=r²

Equation of required circle

=>(x-a)²+(y-a)²={√{(a-h)²+(a-k)²}}²

=>(x-a)²+(y-a)²=(a-h)²+(a-k)²

=>x²+a²-2ax+y²+a²-2ay=a²+h²-2ah+a²+

k² -2ak

=>x²+y²-2ax-2ay=h²-2ah+k²-2ak

=>x²+y²-2ax-2ay+2ah+2ak-h²-k²=0

=>x²+y²-2a(x+y)+2a(h+k)-(h²+k²)=0