MCQs (One or More Correct Option): Circle | JEE Advanced - Free MCQ Practice

The given circle is x² + y² – 2rx – 2hy + h² = 0 with centre (r, h) and radius = r.
Clearly circle touches y-axis so one of its tangent is x = 0.

Let y =  mx be the other tangent through origin.
Then length of perpendicular from C (r, h) to y = mx should be equal to r.

⇒ m²r² – 2mrh + h² = m²r² + r²

∴ Other tangent is y = 

or (h² – r²) x – 2rhy = 0

x² + y² = 4 (given) Centre C1 ≡ (0, 0) and R₁ = 2.
Also for circle x² + y² – 6x – 8y – 24 = 0 C₂ ≡ (3, 4) and R₂ = 7.
Again C₁ C₂ = 5 = R₂ – R₁

Therefore, the given circles touch internally such that they can have just one common tangent at the point of contact.

Putting y = c²/x in x² + y² = a²,
we obtain x² + c⁴/x² = a²

⇒ x⁴ – a²x² + c⁴ = 0 … (1)
As x₁, x₂, x₃ and x₄ are roots of (1),
⇒ x₁ + x₂ + x₃ + x₄ = 0 and x₁ x₂ x₃ x₄ = c₄
Similarly, forming equation in y,
we get y₁ + y₂ + y₃ + y₄ = 0 and y₁ y₂ y₃ y₄ = c₄.

There can be two possibilites for the given circle as shown in the figure

∴  The equations of circles can be (x – 3)² + (y – 4)² = 4²
or  (x – 3)² + (y + 4)² = 4²
i.e.  x² + y² – 6x – 8y + 9 = 0
or   x² + y² – 6x + 8y + 9 = 0

Let the equation of circle be x² + y² + 2gx + 2 f y + c = 0
It passes through (0, 1)
∴ 1 + 2f + c = 0 ...(i)
This circle is orthogonal to (x – 1)² + y2 = 16
i.e. x² + y² – 2x – 15 = 0 and x² + y² – 1 = 0
∴ We should have 2g (– 1) + 2f (0) = c – 15 or 2g + c – 15 = 0 ...(ii)
and 2g(0) + 2f (0) = c – 1 or c = 1 ...(iii)
Solving (i), (ii) and (iii), we get c = 1, g  = 7, f = – 1
∴ Required circle is x² + y² + 14x – 2y + 1 = 0
With centre (– 7, 1) and radius = 7
∴ (b) and (c) are correct options.

Circle : x² + y² = 1
Equation of tangent at P(cosθ, sinθ) x cosθ + y sinθ = 1 ...(1)
Equation of normal at P y = x tanθ ...(2)
Equation of tangent at S is x = 1

∴ Equation of line through Q and parallel to RS is

∴ Intersection point E of normal and

 = x tanθ ⇒ 

∴ Locus of E   or y² = 1 – 2x

It is satisfied by the points  and