Test: Comprehension Based Questions: Circle - JEE MCQ

Without loss of generality we can assume the square ABCD with its vertices A (1, 1), B  (–1, 1), C (–1, –1), D (1, –1)
P to be the point (0, 1) and Q as (,0 ).

Then, 

Let C' be the said circle touching C₁ and L, so that C₁ and C' are on the same side of L. Let us draw a line T parallel to L at a distance equal to the radius of circle C₁, on opposite side of L.
Then the centre of C' is equidistant from the centre of C₁ and from line T.
⇒ locus of centre of C' is a parabola.

Since S is equidistant form A and line BD, it traces a parabola. Clearly, AC is the axis, A (1, 1) is the focus

and  is the vertex of parabola.

T₂ T₃ = latus rectum of parabola

∴ Area (ΔT₁T₂T₃) =

Slope of CD = 

∴ Parametric equation of CD is

∴ Two possible coordinates of C are

 or 

As (0, 0) and C lie on the same side of PQ

  should be the coordinates of C.
NOTE THIS STEP :  Remember (x₁, y₁) and (x₂, y₂) lie on the same or opposite side of a line ax + by + c = 0 according as 

∴ Equation of the circle is

ΔPQR is an equilateral triangle, the incentre  C must coincide with centriod of ΔPQR and D, E, F must concide with the mid points of sides PQ, QR and RP respectively.

Also  

Writing the equation of side PQ in symmetric form we

get, 

∴ Coordinates of P = 

 and 

coordinates of  

Let coordinates of R be (α,β) , then using the formula for centriod of Δ we get

⇒ α = 0 and β = 0

∴ Coordinates of R = (0, 0)

Now coordinates of E = mid point of QR  =

and coordinates of F = mid point of PR = 

Equation of side QR is y = and equation of side RP is  y = 0

Paragraph 3 Given the implicit function y3 – 3y + x = 0

For x ∈(–∞, –2) ∪ (2,∞) it is y = f (x) real valued differentiable function and for x ∈ (–2, 2) it is y = g(x) real valued differentiable function.

Equation of tangent PT to the circle x² + y² = 4 at the point  

Let the line L, perpendicular to tangent PT be

As it is tangent to the circle (x – 3)² + y² = 1

∴ length of perpendicular from centre of circle to the tangent = radius of circle.

= – 1 or  – 5

∴ Equation of L can be 

From the figure it is clear that the intersection point of two direct common tangents lies on x-axis.
Also ΔPT₁C₁ ~ ΔPT₂C₂ ⇒ PC₁ : PC₂ = 2 : 1
or P divides C₁C₂ in the ratio 2 : 1 externally
∴ Coordinates of P are (6, 0) Let the equation of tangent through P be y = m (x – 6)
As it touches x² + y² = 4

36 m² = 4(m² + 1)

∴ Equations of common tangents are

Also x = 2 is the common tangent to the two circles.