Test: Conic Sections - 1 - JEE MCQ

x² + y² + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)² + (3)² - 5} 
= √{9 + 9 - 5} 
= √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
=> -3*2 - 3 + λ = 0
=> -6 - 3 + λ = 0
=> -9 + λ = 0
=> λ = 9

Given equation of circles are 

S₁ = x² + y² + 2x + 6y = 0 ..................1

S₂ = x² + y² − 4x − 2y − 6 = 0 ............2

Subtract equation 1 - equation 2, we get

S₁ - S₂ = 6x + 8y + 6 = 0

=> 6x + 8y + 6 = 0 ............3

Cente of the circle S₂ is (2, 1)

Now, the length of the perpendicular from the center (2, 1) of of the circle 2 upon the common chord 3 is

l = (6*2 + 8*1 + 6)/√{6² + 8² }

=> l = (12 + 8 + 6)/√{36 + 64}

=> l = 26/√100

=> l = 26/10

=> l = 13/5

Radius of the circle 2 is

r = √{2² + 1² - (-6)}

=> r = √{4 + 1 + 6}

=> r = √11

Now length of common chord = 2√{r² - l² }

 = 2√{(√11)² - (13/5)² }

= 2√{11 - 169/25}

= 2√{(275 - 169)/25}

= 2√{106/25} unit

Given equation of circles are
x²+y²+6x+6y=0....(i)
and x²+y²−12x−12y=0....(ii)
Here, g1 = 3,f2 = 3, g2 = −6 and f₂ = −6
∴ Centres of circles are C₁(−3,−3) and C₂(6,6) respectively and radii are r₁ = 3√2 and r₂ = 6√2 respectively.
Now, C₁C₂ = √[(6+3)² + (6+3)²]
= 9√2
and r₁ + r₂
​= 3(2)^1/2 + 6(2)^1/2
= 9(2)^1/2
​⇒ C₁C₂ = r₁ + r₂
∴ Both circles touch each other externally.

The standard equation of a circle is given by:

(x-a)² + (y-b)² = c²

Where (a,b) is the coordinates of center of the circle and c is the radius. 
if c become zero then the radius of circle become zero then there will be no circle exist hence the radius can not be zero. So the correct option is D
 

ax²+by²+2hxy+2gx+2fy+c = 0
In the circle equation,
Coefficients of x² & y² are the same and non-zero.
Therefore, a = b ≠ 0
There is no term for xy. Therefore, h = 0.
Radius of the circle: r² = g² + f² – c,
So, g² + f² – c > 0

Length of tangent from an external point (h,k) to  S = √(S(h,k))
S(2,-3) = √[2(2)² + 2(-3)² - 1] 
=> S(2,-3) = √[8 + 18 - 1]  
=> S(2,-3) = √25
=> S(2,-3) = 5

 Parabola is x² – 8x + 2y + 7 = 0 
∴   (x – 4)² = – 2y – 7 + 16 
∴   (x – 4)² = – 2[y – (9/2)]
 ∴   x² = – 4ay  
 ⇒ x = x – 4, y = y – (9/2) and 2 = 4a
 i.e. a = (1/2) 
 Its focus is given by x = 0 and y = 0 i.e. x – 4 = 0   and     y – (9/3) = 0 
∴    x = 4    and y = (9/2) 
∴ focus [4, (9/2)].