Test: Circle- 1 - JEE MCQ

The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .
Now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .
we also know, General equation of circle is
x² + y² + 2gx + 2fy + C = 0
when point (0,0)
(0)² + (0)² + 2g(0) + 2f(0) + C = 0
0 + 0 + 0 + 0 + C = 0
C = 0 -------(1)
when point (a,0)
(a)² + (0)² + 2g(a) + 2f(0) + C = 0
a² + 2ag + C = 0
from equation (1)
a² + 2ag = 0
a(a + 2g) = 0
g = -a/2
when point ( 0, b)
(0)² + (b)² + 2g(0) + 2f(b) + C = 0
b² + 2fb + C = 0
f = -b/2
Now, equation of circle is
x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }
x² + y² - ax - by = 0
As we know that, a=2, b=4
x^2 + y^2 - 2x - 4y = 0

x²+y²+4x-6y=5
Circle Equation
(x-a)²+(y-b)²=r² is the circle equation with a radius r, centered at (a,b)
Rewrite x²+y²4x-6y=5 in the form of circle standard circle equation
(x-(-2))²+(y-3)²=(3√2)^{2
Therefore the circle properties are:}
(a,b) = (-2,3), r = 3√2

 Radius of circle is given by -
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)²+(y-k)²=r²
equation is (x-2)² + (y-1)² = (√37)²
x² + 4 - 4x + y² + 1 - 2y = 37
x² + y² - 4x - 2y - 32 = 0

If the circle lies in second quadrant
The equation of a circle touches both the coordinate axes and has radius a is
x² + y² + 2ax - 2ay + a² = 0
Radius of circle, a = 5
x² + y² + 10x - 10y + 25 = 0 

Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the centre of the circle passes through (0, 0),
(0 – h)² + (0 – k)² = r²
⇒ h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k².
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)² + (0 – k)² = h² + k² …………… (1)
(0 – h)² + (b – k)² = h² + k² ………… (2)
From equation (1), we obtain a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.
From equation (2), we obtain h² + b² – 2bk + k² = h² + k²
⇒ b² – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or(b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =b/2. Thus, the equation of the required circle is

Equation of any line through (3,11) is 
Then the coordinates of a point on this line at a distance r from (3,11) are
(3+rcosθ,11+rsinθ)
and if this lies on the given circle
x² + y² = 9 then (3 + rcosθ)² + (11 + rsinθ)² = 9
⇒ 9 + 121 + 2r(3cosθ + 11sinθ) + r² = 9
⇒ r² + 2r(3cosθ + 11sinθ) + 121 = 0
which is quadratic in r, gives two values of r say r₁ and r₂ and hence the distances of the points A and B from P. Thus, PA.PB = r₁r₂ = 121

X² +Y² =25    Points =(-2,-5)
Put the points in the variables
(-2)² + (-5)² = 29
As 29 > 25  (lies outside).

The line y = x meets the circle x² + y² − 2x = 0 when 2x² − 2x = 0
⇒x=0 or 1.
Hence, the equation of the required circle is x(x−1)+y(y−1)=0