Area of a circle is 81π and the equations of the normal to the circle

1.	Area of a circle is 81π and the equations of the normal to the circle are 2y + 3x - 5 = 0 and 2y - 3x + 5 = 0. Find the equation of the circle.1. (x - \(\frac{5}{3}\))2 + y2 = 92. (x - \(\frac{5}{3}\))2 + y2 = 813. (x + \(\frac{5}{3}\))2 + y2 = 814. (x + \(\frac{5}{3}\))2 + y2 = 9
Answer» Correct Answer - Option 2 : (x - \(\frac{5}{3}\))² + y² = 81 Concept: Standard equation of a circle: \(\rm (x-h)^2+(y-k)^2=R^2\) Where centre is (h, k) and radius is R. Note: The normal to the circle passes through the centre of the circle. Distance between a point on a circle and the centre is the radius of the circle. Distance between 2 points (x1, y1) and (x2, y2) is: D = \(\rm \sqrt{(y_2-y_1)^2 + (x_2 -x_1)^2}\) Area of the circle = πr² Calculation: Given normal to the circle are : 2y - 3x + 5 = 0 ...(i) 2y + 3x - 5 = 0 ...(ii) Substracting (ii) from (i) -6x + 10 = 0 ⇒ x = \(\boldsymbol{5\over3}\) From equation (ii) 2y + 3(\({5\over3}\)) - 5 = 0 ⇒ y = 0 ∴ Coordinates of the centre are (\(\boldsymbol{5\over3}\), 0) Given the area of circle = 81π πr² = 81π ⇒ r2 = 81 ∴ The equation of circle is: \(\rm (x-h)^2+(y-k)^2=R^2\) ⇒ (x - \({5\over3}\))2 + (y - 0)2 = 81 ⇒ (x - \(\boldsymbol{5\over3}\))2 + y2 = 81

Area of a circle is 81π and the equations of the normal to the circle are 2y + 3x - 5 = 0 and 2y - 3x + 5 = 0. Find the equation of the circle.1. (x - \(\frac{5}{3}\))2 + y2 = 92. (x - \(\frac{5}{3}\))2 + y2 = 813. (x + \(\frac{5}{3}\))2 + y2 = 814. (x + \(\frac{5}{3}\))2 + y2 = 9

Answer» Correct Answer - Option 2 : (x - \(\frac{5}{3}\))² + y² = 81

Concept:

Standard equation of a circle:

\(\rm (x-h)^2+(y-k)^2=R^2\)

Where centre is (h, k) and radius is R.

Note: The normal to the circle passes through the centre of the circle.

Distance between a point on a circle and the centre is the radius of the circle.

Distance between 2 points (x1, y1) and (x2, y2) is:

D = \(\rm \sqrt{(y_2-y_1)^2 + (x_2 -x_1)^2}\)

Area of the circle = πr²

Calculation:

Given normal to the circle are :

2y - 3x + 5 = 0 ...(i)

2y + 3x - 5 = 0 ...(ii)

Substracting (ii) from (i)

-6x + 10 = 0

⇒ x = \(\boldsymbol{5\over3}\)

From equation (ii)

2y + 3(\({5\over3}\)) - 5 = 0

⇒ y = 0

∴ Coordinates of the centre are (\(\boldsymbol{5\over3}\), 0)

Given the area of circle = 81π

πr² = 81π

⇒ r2 = 81

∴ The equation of circle is:

\(\rm (x-h)^2+(y-k)^2=R^2\)

⇒ (x - \({5\over3}\))2 + (y - 0)2 = 81

⇒ (x - \(\boldsymbol{5\over3}\))2 + y2 = 81