Question

What is the equation of the circle with center (-2,5) and radius √3

Answer 1

Answer:

x^2 + y^2 + 4x - 10y + 26 = 0

Step-by-step explanation:

( x , y ) be any point of the the circle.

Equation of the circle is :

⇒ √[ { x - ( - 2 ) }^2 + { y - 5 }^2 ] = √3

⇒  ( x + 2 )^2 + ( y - 5 )^2 = ( √3 )^2

⇒ x^2 + 2^2 + 2( 2 )( x ) + y^2 + 5^2 - 2( 5 )( y ) = 3

⇒ x^2 + 4 + 4x + y^2 + 25 - 10y = 3

⇒ x^2 + y^2 + 4x - 10y + 29 - 3 = 0

⇒ x^2 + y^2 + 4x - 10y + 26 = 0

Hence the required equation is x^2 + y^2 + 4x - 10y + 26 = 0 .

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

What is the equation of the circle with center (-2,5) and radius √3

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

Let,

• center of circle is O(x,y) = O(-2,5)
• Radius(r) = √3

$\Large{\underline{\mathfrak{\bf{\pink{Find}}}}}$

• Equation of circle

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

we know,

$\Large{\underline{\mathfrak{\bf{\pink{Equation\:of\:circle}}}}}$

$\small\boxed{\sf{\red{\:\sqrt{(x-a)^2+(y-b)^2}\:=\:r}}}$

• where,point of (a,b) is centre
• r = radius

Here, O(-2,5) is centre and √3 is radius

Then,

$\mapsto\sf{\:\sqrt{(x-(-2))^2+(y-5)^2}\:=\:(\sqrt{3})} \\ \\ \small\sf{\green{\:\:squaring\:both\:side}} \\ \\ \mapsto\sf{\:(x+2)^2+(y-5)^2\:=\:3} \\ \\ \mapsto\sf{\:(x^2+4+2.2.x)+(y^2+25-2.5.y)\:=\:3} \\ \\ \mapsto\sf{\:(x^2+y^2+4x-10y+29-3)\:=\:0} \\ \\ \mapsto\sf{\pink{\:x^2+y^2+4x-10y+26\:=\:0\:\:\:Ans.}}$