Conic Sections Test - 1

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
x ²+ y ²+ 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
a ²+ 2ag + C = 0
from equation (1)
a ²+ 2ag = 0
a(a + 2g) = 0
g = -a/2
when point ( 0, b)
(0)²+ (b)²+ 2g(0) + 2f(b) + C = 0
b ²+ 2fb + C = 0
f = -b/2
Now, equation of circle is
x ²+ y ²+ 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }
x ²+ y ²- ax - by = 0
As we know that, a=2, b=4
x^2 + y^2 - 2x - 4y = 0

The equation of a circle with center (h,k) and radius r is given by (x −h)² +(y −k)² =r² . 
For a circle centered at the origin, this becomes the more familiar equation x² +y² =r²

A circle is the set of points a fixed distance from a center point, 

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 +y² 4x-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 = √(-1 ²+ 6 ²)
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

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