Question

All formulas of " Circle "

class 11th as per JEE syllabus

Answer 1

The equation of a circle with its center at C(x0, y0) and radius r is: (x – x0)2 + (y – y0)2 = r2   

If x0 = y0 = 0 (i.e. the centre of the circle is at origin) then equation of the circle reduce to x2 + y2 = r2.

If r = 0 then the circle represents a point or a point circle.

The equation x2 + y2 + 2gx + 2fy + c = 0 is the general equation of a circle with centre (–g, –f) and radius √(g2+f2-c).

Equation of the circle with points P(x1, y1) and Q(x2, y2) as extremities of a diameter is (x – x1) (x – x2) + (y – y1)(y – y2) = 0. 

For general circle, the equation of the chord is x1x + y1y + g(x1 + x) + f(y1 +y) + c = 0

For circle x2 + y2 = a2, the equation of the chord is x1x + y1y = a2 

The equation of the chord AB

(A ≡ (R cos α, R sin α); B ≡ (R cos β, R sin β)) of the circle x2 + y2 = R2 is given by x cos ((α + β )/2) + y sin ((α - β )/2) = a cos ((α - β )/2) 

The point P(x1, y1) lies outside, on, or inside a circle  ?

           S ≡ x2 + y2 + 2gx + 2fy + c = 0, according as S1 ≡ x12 + y12 + 2gx1 + 2fy1 + c > = or < 0.

The equation of the chord of the circle x2 + y2 + 2gx + 2fy +c=0 with M(x1, y1) as the midpoint of the chord is given by:  

           xx1 + yy1 + g(x + x1) + f(y + y1) = x12 + y12 + 2gx1 + 2fy1  i.e. T = S1 

Answer 2

Numerical: Find the equation of the circle with centre (–3, 2) and radius 4.

Solution Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is

(x + 3)2 + (y –2)2 = 16

 

Numerical: Find the centre and the radius of the circle x2 + y2 + 8x + 10y – 8 = 0

Solution The given equation is (x2 + 8x) + (y2 + 10y) = 8

Completing the squares, we get

(x2 + 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25

i.e. (x + 4)2 + (y + 5)2 = 49 = 72

Therefore, the given circle has centre at (– 4, –5) and radius 7.

I DON'T KNOW FORMULAE OF CIRCLE WITH CORRECTNESS SO I GIVE YOU SOME NUMERICAL PLEASE DON'T DELETE MY ANSWER.