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Mathematics Question on Coordinate Geometry

Let a circle passing through (2, 0) have its centre at the point (h,k)(h, k). Let (xc,yc)(x_c, y_c) be the point of intersection of the lines 3x+5y=13x + 5y = 1 and (2+c)x+5c2y=1(2 + c)x + 5c^2y = 1. If h=lim⁡c→1xch = \lim_{c \to 1} x_c and k=lim⁡c→1yck = \lim_{c \to 1} y_c, then the equation of the circle is:

A

25x2+25y2−20x+2y−60=025x^2 + 25y^2 - 20x + 2y - 60 = 0

B

5x2+5y2−4x−2y−12=05x^2 + 5y^2 - 4x - 2y - 12 = 0

C

25x2+25y2−2x+2y−60=025x^2 + 25y^2 - 2x + 2y - 60 = 0

D

5x2+5y2−4x+2y−12=05x^2 + 5y^2 - 4x + 2y - 12 = 0

Answer

25x2+25y2−20x+2y−60=025x^2 + 25y^2 - 20x + 2y - 60 = 0

Explanation

Solution

Let the circle pass through the point (2,0)(2, 0) with its center at (h,k)(h, k). The point of intersection of the lines 3x+5y=1and(2+c)x+5c2y=13x + 5y = 1 \quad \text{and} \quad (2 + c)x + 5c^2y = 1 is given as (xc,yc)(x_c, y_c), where: x=1−c22+c−3c2,y=1−3c5(2+c−3c2).x = \frac{1 - c^2}{2 + c - 3c^2}, \quad y = \frac{1 - 3c}{5(2 + c - 3c^2)}.

Step 1: Limits of hh and kk

h=lim⁡c→1x=lim⁡c→1(1−c)(1+c)(1−c)(2+3c)=25,h = \lim_{c \to 1} x = \lim_{c \to 1} \frac{(1 - c)(1 + c)}{(1 - c)(2 + 3c)} = \frac{2}{5},
k=lim⁡c→1y=lim⁡c→1c−1−5(c−1)(3c+2)=−125.k = \lim_{c \to 1} y = \lim_{c \to 1} \frac{c - 1}{-5(c - 1)(3c + 2)} = -\frac{1}{25}.
Thus, the center of the circle is: (25,−125).\left( \frac{2}{5}, -\frac{1}{25} \right).

Step 2: Radius of the circle

Using the distance formula, the radius is:
r=(2−25)2+(0−(−125))2.r = \sqrt{\left( 2 - \frac{2}{5} \right)^2 + \left( 0 - \left(-\frac{1}{25}\right)\right)^2}.
Simplify:
r=(105−25)2+(125)2=(85)2+(125)2.r = \sqrt{\left(\frac{10}{5} - \frac{2}{5}\right)^2 + \left(\frac{1}{25}\right)^2} = \sqrt{\left(\frac{8}{5}\right)^2 + \left(\frac{1}{25}\right)^2}.
r=6425+1625=1600+1625=1601625=160125.r = \sqrt{\frac{64}{25} + \frac{1}{625}} = \sqrt{\frac{1600 + 1}{625}} = \sqrt{\frac{1601}{625}} = \frac{\sqrt{1601}}{25}.

Step 3: Equation of the circle

The general equation of the circle is:
(x−25)2+(y+125)2=(160125)2.\left(x - \frac{2}{5}\right)^2 + \left(y + \frac{1}{25}\right)^2 = \left(\frac{\sqrt{1601}}{25}\right)^2.
Simplify:
(x−25)2+(y+125)2=1601625.\left(x - \frac{2}{5}\right)^2 + \left(y + \frac{1}{25}\right)^2 = \frac{1601}{625}.
Multiply through by 625:
25(x−25)2+25(y+125)2=1601.25\left(x - \frac{2}{5}\right)^2 + 25\left(y + \frac{1}{25}\right)^2 = 1601.
Expand:
25x2−20x+4+25y2+2y+125=1601.25x^2 - 20x + 4 + 25y^2 + 2y + \frac{1}{25} = 1601.
Simplify:
25x2+25y2−20x+2y−60=0.25x^2 + 25y^2 - 20x + 2y - 60 = 0.

Final Answer:

25x2+25y2−20x+2y−60=0.\boxed{25x^2 + 25y^2 - 20x + 2y - 60 = 0.}