Question -
Answer -
(i) (x – 1)2 +y2 = 4
Given:
The equation (x – 1)2 +y2 = 4
We need to find thecentre and the radius.
By using the standardequation formula,
(x – a)2 +(y – b)2 = r2 …. (1)
Now let us convertgiven circle’s equation into the standard form.
(x – 1)2 +y2 = 4
(x – 1)2 +(y – 0)2 = 22 ….. (2)
By comparing equation(2) with (1), we get
Centre = (1, 0) andradius = 2
∴ The centre of thecircle is (1, 0) and the radius is 2.
(ii) (x + 5)2 +(y + 1)2 = 9
Given:
The equation (x + 5)2 +(y + 1)2 = 9
We need to find thecentre and the radius.
By using the standardequation formula,
(x – a)2 +(y – b)2 = r2 …. (1)
Now let us convertgiven circle’s equation into the standard form.
(x + 5)2 +(y + 1)2 = 9
(x – (-5))2 +(y – ( – 1))2 = 32 …. (2)
By comparing equation(2) with (1), we get
Centre = (-5, -1) andradius = 3
∴ The centre of thecircle is (-5, -1) and the radius is 3.
(iii) x2 + y2 –4x + 6y = 5
Given:
The equation x2 +y2 – 4x + 6y = 5
We need to find thecentre and the radius.
By using the standardequation formula,
(x – a)2 +(y – b)2 = r2 …. (1)
Now let us convertgiven circle’s equation into the standard form.
x2 + y2 –4x + 6y = 5
(x2 –4x + 4) + (y2 + 6y + 9) = 5 + 4 + 9
(x – 2)2 +(y + 3)2 = 18
(x – 2)2 +(y – (-3))2 = (3√2)2 … (2)
By comparing equation(2) with (1), we get
Centre = (2, -3) andradius = 3√2
∴ The centre of thecircle is (2, -3) and the radius is 3√2.
(iv) x2 + y2 –x + 2y – 3 = 0
Given:
The equation x2 +y2 – x + 2y – 3 = 0
We need to find thecentre and the radius.
By using the standardequation formula,
(x – a)2 +(y – b)2 = r2 …. (1)
Now let us convertgiven circle’s equation into the standard form.
x2 + y2 –x + 2y – 3 = 0
(x2 –x + ¼) + (y2 + 2y + 1) – 3 – ¼ – 1 = 0
(x – ½)2 +(y + 1)2 = 17/4 …. (2)
By comparing equation(2) with (1), we get
Centre = (½, – 1) andradius = √17/2
∴ The centre of thecircle is (½, -1) and the radius is √17/2.