(A) Substitute y = x in the circle equation. Therefore
2x2 - 2x = 0
⇒ x = 0,1
Hence, A(0, 0) and B = (1, 1). Therefore, the equation of the circle with AB as diameter is given by
x(x - 1) + y(y - 1) = 0
⇒x2 + y2 - x - y = 0
Answer: (A) → (*)
(B) The equation of the tangent at (5, 5) to the given circle is
x(5) + y(5) -(x + 5) - 2(y + 5) - 20 = 0
That is, 4x + 3y - 35 = 0. The required circle is of the form x2 + y2 - 2x - 4y - 20 + λ (4x + 3y - 35) = 0. The radius of the circle is 5. This implies that
λ = 0 gives the circle x2 + y2 - 2x - 4y + 20 = 0 and
λ = -4 gives the required circle which is x2 + y2 - 18x - 16y + 120 = 0.
Aliter: Since the radius of the circle x2 + y2 - 2x - 4y - 20 = 0 is 5 and the radius of the required circle is also 5, it follows that the contact must be external contact and the point (5, 5) must be the midpoint of the segment joining the centres. Hence, if (h, k) is the centre of the required circle, we have
1 + h/2 = 5 and 2 + k/2 = 5
⇒ h = 9,k = 8
Hence, the equation of the required circle is
(x - 9)2 + (y - 8)2 = 25
⇒ x2 + y2 - 18x - 16y + 120 = 0
Answer: (B)→(p)
(C) We have the equation of the two lines as
3x + 5y = 1
and (2 + c)x + 5c2y = 1
On solving these equations, we get
Hence, the centre of the circle = (2/5, -1/25). Since the circle passes through (2, 0), its radius is
Therefore, equation of the required circle is
That is,
25(x2 + y2)- 20x + 2y - 60 = 0
Answer: (C) → (q)
(D) Let ≡* x2 + y2 - 4x - 2y - 8 = 0
and ' ≡* x2 + y2 - 2x - 4y - 8 = 0
Let L ≡ - ' ≡ -2x + 2y = 0
Hence, the required circle is of the form
*S +λ L≡ x2 + y2 - 4x - 2y - 8 +*λ(x - y)= 0
This passes through (-1, -4). This implies
1 + 16 + 4 - 8 + λ(-1 -4) = λ = 1
Therefore, the required circle is
x2 + y2 - 3x - 3y - 8 = 0
Answer: (D) → (r)
