Let a circle passing through (2,0) have its centre at the point (h, k). Let (x_c, y_c) be the point of intersection of the lines 3 x+5 y=1 and (2+c) x+5 c² y=1. If h=lim c → 1 x_c and k=lim c → 1 y_c, then the equation of the circle is :

Solution

Maths sir · 5 steps

At c = 1 both lines become the same line, so solve for general c, cancel (1 − c), then take the limit.

1. Why a limit. Put c = 1 in the second line and you get 3x + 5y = 1 again. The two lines coincide, so there is no single intersection point at c = 1. Solve for c ≠ 1 first.
2. Eliminate y. From the first line, 5y = 1 - 3x. Substitute into the second: (2+c)x + c²(1 - 3x) = 1, so x(2 + c - 3c²) = 1 - c². Both sides carry a factor (1 - c).

   x_c = (1-c)(1+c)/(1-c)(3c+2) = (1+c)/(3c+2)

3. Take the limit. As c → 1, h = 2/5. Then k = (1 - 3h)/5 = (1 - 6/5)/5 = -1/25.
4. Build the circle. Write it as x² + y² - 2hx - 2ky + d = 0, which is x² + y² - 4/5x + 2/25y + d = 0. Passing through (2, 0) gives 4 - 8/5 + d = 0, so d = -12/5.
5. So. Multiplying by 25 gives 25x² + 25y² - 20x + 2y - 60 = 0, which is option (b).

The idea to keep. When a parameter makes two lines coincide, solve for the general value, cancel the common factor, and only then substitute. That is the same move you use for any 0/0 limit.

Answer from the NTA final answer key. Solution by PrepWiser.