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.

Given x₁, x₂ are the roots of the equation.
x² + 2x − 3 = 0
⇒ x² + 3x − x − 3 = 0
⇒x(x+3)−1(x+3)=0
⇒(x−1)(x+3)=0
⇒ x₁ = −3, x₂ = 1
And y₁,y₂ are the roots of the equation.
y²+4y−12=0
⇒y²+6y−2y−12=0
⇒y(y+6)−2(y+6)=0
⇒(y−2)(y+6)=0
⇒y₁ = −6,
y₂ = 2
Points are P(−3,−6) &Q(1,2)
Since Pand Q are the endpoints of a diameter.
∴ Centre = Mid point of PQ

=(−1,) −2

In an equilateral triangle, the circumcentre of a circle lies on the centroid of the triangle and centroid divides median in 2:1.
Hence, OC=2/3×(length of median)
⇒ OC = 2/3 × 3a = 2a
Here, radius of circle is 2a

∴ Required equation of circle is
x²+y²=4a²

Equation of common chord is
S₁ - S₂ 
⇒ 6y + 6 = 0 ⇒ y = -1
Putting y=−1 in Eq. (i), we get
∴ x²+1+2x−3+2=0
x²+2x=0⇒x=0,−2
∴ End points of diameter are
(0,−1) and (−2,−1)
Equation of circle is
(x−0) (x+2)+(y+1)(y+1) = 0
⇒ x²+2x+y²+2y+1 = 0

The point A is (acosα,2+asinα)
since the triangle ABC is equilateral its circum centre (0,2) coincides with its centroid

x²+y²−4x−6y−3=0
⇒  (x−2)²+(y−3)²=42
Parametric forms of circle will be
x−2=4cosθ and y−3=4sinθ
⇒  x=2+4cosθ and y=3+4sinθ
Now, x + y = 5 + 4(sinθ + cosθ), − √2 ≤ sinθ + cosθ ≤√2 for maximum value of , value of (x+y) will be minimum, so (x + y) min = 5 + 4(−√2) = 5−4√2
Hence, answer is 

Let (h,k) be the centre of the circle
Now AO=CO⇒AO²=CO²
⇒(2−h)²+(3−k)² 
=(3+h)²+(2−k)²
⇒k=−5h...(i)

and AO=BO⇒AO²=BO² 
⇒(2−h)²+(3−k)²=(3−h)²+(1+k)² 
⇒2h−8k=−3...(ii)
after solving (i) \& (ii), we get

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

Given,
(x−2009)²+(x−2009)²=16
We have to find the number of points with integral coordinates which are interior to the circle. for this 
Number of integral (coordinate) solution = Number of points,
(x)²+(y)² <(4)² where x∈[−3,3] and y∈[−3,3]
clearly x and y∈{−3,−2,−1,0,1,2,3}
The ordered pairs of (x, y)={−3,−2,−1,0,1,2,3}×{−3,−2,−1,0,1,2,3}
⇒ The odered pairs
={(−3,−3),(−2,−3),(−1,−3),(0,−3),(1,−3),(2,−3),(3,−3)(−3,−2),(−2,−2),(−1,−2),(0,−2),(1,−2),(2,−2),(3,−2)  .(−3 ,3) , (−2, 3),  (−1, 3),   (0, 3),   (1, 3),  (2, 3), (3, 3)}
The total number of ordered pairs =49
Out of these 49 ordered pairs, 4 ordered  {(−3,−3),(−3,3),(3,−3),(3,3)} pairs doesn't satisfy.

⇒ The number of points with integral coordinates which are interior to the  given circle =49−4=45
⇒ By shifting the origin to (2009,−2009), the number of points with integral coordinates will remain the same.  

Let, point M=(h,k), then

We observe that, (x−1)(x−3)+(y−2)(y−4)=0, is equation of circle in diametric form. Hence, AB is a diameter and (x,y) lies on the circle.
Diameter AB=2√2
So, radius = √2
Area(∆ABC)=3

Height h=3√2> Radius
Hence, no such C exists.

Here radius of smaller circle, 
Clearly, from the figure the radius of bigger circle

Let the centre of the required circle be C(h,h).
Centre of the given circle B(2,2) and radius is 2
We have, ∠COD =∠CBE  = π/4
CB=h+2
and BE=h−2

Let the equation of circle be x²+y²+2gx+2fy+c=0.
Then the radius, 
Centre: (−g,−f).
But since circle touches Y-axis, the absolute value of abscissa is equal to the radius, |−g|=r.
⇒g²=r²⇒g²=g²+f²−c
⇒c=f²
Centre of the circle, x²+y²−6x−6y+14=0: (3,3)

As the two circles touch externally, the distance between
centres is equal to the sum of radii. 

Hence, locus of the centre (−g,−f) obtained by substituting −g=x, −f=y⇒g=−x, f=−y, is y²−10x−6y+14=0.

Given circle x²+y²−2x=0
Centre and radius of the given circle are (1,0) and 1.
If we take an image of a circle then radius will be the same, only center coordinate will change.
Let the center of the image circle is  (x₁,y₁) Hence, (x₁, y₁) be the image of the point (1,0) w.r.t. the line x+y=2, then using image theorem,

Distance from centre (2, 1) to the line 3x+4y−5= radius of circle

Equation of circle is

Given circles are x² + y² − 2x + 8y + 13 = 0
and x² + y² − 4x + 6y + 11 = 0.
Here, C₁ = (1,−4),C₂ = (2,−3),

AB=2 AM
AB²=4 AM² 

n=1 or n=2 (∵ length should be +ve)
Hence required sum
=2(8−1²+8−2²)
=2×11=22

The coordinates of the centre and radius of the given circle are (1, 1) and 2 respectively. Let AB be the chord subtending an angle of 120° at the center. Let M be the midpoint of AB and let its coordinates be (h, k).
In ∆OAM,

The equation of circle is x² + y² = 16
Centre, C₁(0,0); Radius, r₁=4
Equation of circle:  x²+y²+2y=0
Centre C₂ (0, −1); radius, r₂=1

∴ The sum of distances between the centres and smallest radius is less than the bigger radius.
Hence, second circle will completely lie inside the first circle.
Therefore, no common tangents.