Question

Find the centre and radius of the circle x2 + y2 + 2x + 4y - 5 = 0​

Answer 1

Answer:

Explanation:

The standard eqn of a circle with centre (a,b) and radius r

is: (x−a)2+(y−b)2=r2

for  x2+y2−2x+4y−4=0 we need to complete the square.

x2+y2−2x+4y−4=0

x2−2x+y2+4y−4=0

(x2−2x+12)+(y2+4y+22)−12−22−4=0

(x−1)2+(y+2)2−9=0

(x−1)2+(y+2)2=9

centre (1,−2)

radius

Step-by-step explanation:

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Answer 2

• The center of the circle is ( - 1, - 2)

• The radius of the circle is √10 unit

Given :

The equation of a circle is x² + y² + 2x + 4y - 5 = 0

To find :

The centre and radius of the circle

Formula :

The equation of any circle with center (h,k) and of radius r is

(x - h)² + (y - k)² = r²

Solution :

Step 1 of 2 :

Write down the given equation of the circle

Here the given equation of the circle is

x² + y² + 2x + 4y - 5 = 0

Step 2 of 2 :

Find centre and radius of the circle

$\displaystyle \sf{ {x}^{2} + {y}^{2} + 2x + 4y - 5 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} + 2x + {y}^{2} +4y - 5 = 0}$

$\displaystyle \sf{ \implies {x}^{2} + 2.x.1 + {1}^{2} + {y}^{2} + 2.y .2+ {2}^{2} - {1}^{2} - {2}^{2} - 5= 0}$

$\displaystyle \sf{ \implies {(x + 1)}^{2} + {(y + 2)}^{2} -1 - 4 - 5= 0}$

$\displaystyle \sf{ \implies {(x + 1)}^{2} + {(y + 2)}^{2} - 10= 0}$

$\displaystyle \sf{ \implies {(x + 1)}^{2} + {(y + 2)}^{2} = 10}$

$\displaystyle \sf{ \implies {(x + 1)}^{2} + {(y + 2)}^{2} = { (\sqrt{10} )}^{2} }$

Which is of the form (x - h)² + (y - k)² = r²

Where h = - 1 , k = - 2 , r = √10

The center of the circle is ( - 1, - 2)

The radius of the circle is √10 unit

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