Test: 35 Year JEE Previous Year Questions: Circle - JEE MCQ

In isosceles triangle side AB = CA For right angled triangle, BC² = AB² + AC²

So, here BC = or BC² = 52

or

Equation of AB is x cosα + y sinα = p;

So co-ordinates of A and B are

So coordinates of  midpoint of AB are

⇒ cosα = p/2x₁ and sinα = p/2y₁  ;

cos²α + sin²α =  1

Locus of (x₁, y₁) is  

Put x = 0 in the given equation ⇒ by² + 2 fy + c = 0.
For unique point of intersection f²– bc = 0 ⇒ af² – abc = 0.
Since  abc + 2fgh – af² –bg² – ch² = 0 ⇒ 2fgh – bg² – ch² = 0

3a + a² – 2 = 0 ⇒ a² + 3a – 2 = 0.;

Co-ordinates of  A = (a cosα , a sinα) Equation of OB,

∴ slope of CA =

Equation of CA

y - a sinα =  (x - a cosα)

⇒ ( y - a sinα) (1 + tan a) = (a cosα - x )(1 - tana)
⇒ (y - a sinα) (cosα+ sinα) = (a cosα- x )(cosα- sinα) 
⇒ y(cos+ sinα) - a sinα cosα - a sin²α = a cos²α - a cosα sinα- x (cosα- sinα)
⇒ y(cosα + sinα) + x(cosα - sinα)= a
y(sinα + cosα ) + x(cosα - sinα ) = a.

Equation of bisectors of second pair of straight lines

is,    qx² + 2 xy - qy²=0 ........(1)

It must be identical to the first pair
x² - 2 pxy -y²=0 ........(2)

from (1) and  (2)

Squaring & adding,

Taking co-ordinates as

Then slope of line joining

and slope of line joining (x, y) and (xr, yr)

∴ m₁ =  m₂

⇒ Points lie on the straight line.

Let the vertex C be (h, k), then the

centroid of Δ ABC is

orIt lies on 2x + 3y  = 1

.

= Locus of  C is 2x + 3y = 9

Let the required line be ......(1)

then a + b = –1 .......(2)

(1) passes through (4, 3) ,

⇒ 4b +3a= ab .................(3)

Eliminating b from (2) and (3), we get a² -4 = 0

⇒ a =±2 ⇒b =-3 or1

∴ Equations of straight lines are

Let the lines be y = m₁x and y = m₂x then

m₁ + m₂ = and  4 m₁m₂ =  

Given m₁ + m₂= 4 m₁m₂

3x + 4y=0 is one of the lines of the pair

6 x² - xy + 4cy²=0 , Put  

we get  

The line passing through the intersection of lines
ax + 2by = 3b=0 and bx - 2ay - 3a=0
is ax + 2by + 3b + λ (bx – 2ay – 3a) = 0
⇒ (a + b λ) x + (2b – 2aλ)y + 3b – 3λa = 0
As this line is parallel to x-axis.
∴ a + bλ = 0  ⇒ λ = – a/b 

⇒ ax + 2by + 3b – (bx – 2ay – 3a) = 0

⇒ ax + 2by + 3b – ax  

So it is 3/2 units below x-axis.

Vertex of tringle is (1, 1) and midpoint of sides through this vertex is (– 1, 2) and (3, 2)

⇒ vertex B and C come out to be (– 3, 3) and (5, 3)

∴ Centroid is 

∵ A is the mid point of PQ , therefore

a = 6, b = 8

∴ Equation of line is or 4x + 3y = 24

Clearly for point P,

a² - 3a<0 and 

Given : The vertices of a right angled triangle A(l, k), B(1, 1) and C(2, 1) and Area of ΔABC = 1 square unit

We know that, area of right angled triangle

⇒ ± (k- 1)=2 ⇒ k = – 1, 3

Given : The coordinates of points P, Q, R are (–1, 0),

(0, 0), respectively..

Slope of QR  = 

⇒ 

Let QM bisects the ∠PQR ,

∴Slope of the line QM = 

∴ Equation of line QM is (y – 0) =  -

Equation of bisectors of lines, xy = 0 are y = ± x

∴ Put y = ± x in the given equation my² + (1 – m²)xy – mx² = 0
∴ mx² + (1 – m²)x² – mx² = 0
⇒ 1 – m² = 0 ⇒ m = ± 1

Slope of  

∴ Slope of perpendicular bisector of  PQ = ( k –1)

Also mid point of

∴ Equation of perpendicular bisector is

⇒ 2y – 7 = 2(k –1) x –(k² –1)

⇒ 2(k – 1)x – 2y + ( 8 – k²) = 0

∴ y-intercept

⇒ 8 – k² = –8  or k² = 16  ⇒ k = ± 4

Let (a², a) be the point of shortest distance on x = y²

Then distance between (a², a) and line x – y + 1 = 0 is given by

It is min when

If the lines p (p² + 1) x – y + q = 0
and (p² + 1)² x + (p² + 1) y +2q = 0
are perpendicular to a common line then these lines must be parallel to each other,

∴ m₁ = m₂

⇒ (p² + 1)  ( p + 1) = 0
⇒ p = – 1
∴ p  can have exactly one value.

Given that

P (1, 0), Q (– 1, 0)  and  

⇒ 3AP = AQ
Let A = (x, y) then 3AP = AQ ⇒ 9 AP²= AQ²
⇒ 9 (x – 1)² + 9y² = (x + 1)² + y²
⇒ 9 x² – 18x + 9 + 9y² = x² +2x +1 + y²
⇒ 8x² – 20x + 8y² + 8 = 0

....(1)

∴ A lies on the circle given by eq (1). As B and C also follow the same condition, they must lie on the same circle.

∴ Centre of circumcircle of Δ ABC  

Slope of line L = 

Slope of line K = 

Line L is parallel to line k.

(13, 32) is a point on L.

Equation of K : y - 4x=3 ⇒ 4x -y + 3=0

Distance between L and K  = 

L₁ : y – x = 0
L₂ : 2x + y = 0
L₃ : y + 2 = 0
On solving the equation of line L₁ and L₂ we get their point of intersection (0, 0) i.e., origin O.
On solving the equation of line L₁ and L₃, we get  P = (– 2, – 2).
Similarly, we get Q = (– 1, – 2) We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [Angle Bisector Theorem of a Triangle]

Let the joining points be A (1,1) and B (2,4).
Let point C divides line AB in the ratio 3 : 2.
So, by section formula we have

Since Line 2x + y = k passes through 

∴ C satisfies the equation 2x + y = k.

Suppose B(0, 1) be an y point on given line and co-ordinate of A is   So, equation of

Reflected Ray is 

From the figure, we have

a = 2, b = , c = 2
x₁ = 0, x₂ = 0, x₃ = 2

Now, x-co-ordinate of incentre is given as

⇒ x-coordinate of incentre =  

Let P, Q, R, be the vertices of ΔPQR

Since PS is the median, S is mid-point of QR

So, 

Now, slope of PS

Since, required line is parallel to PS therefore slope of required line = slope of PS Now, eqn of line passing through (1, –1)  and having slope  is

y-(-1) = 

9y + 9 = –2x + 2 ⇒ 2x + 9y + 7 = 0