Question

If the circle
x²+y²-2gx+6y-19c = 0,g,c∈R
passes through the point (6, 1) and its centre lies on the line x – 2cy = 8, then the length of intercept made by the circle on x-axis is

• A. √11
• B. 4
• C. 3
• D. 2√23

Answer 1

The Correct Option is D

To solve this problem, we first need to understand the equation of the given circle and the conditions provided:

The equation of the circle is:

\(x^2 + y^2 - 2gx + 6y - 19c = 0\)

The center of the circle is at \((g, -3)\). The radius of the circle is:

\(r = \sqrt{g^2 + (-3)^2 + 19c}\)

1. Substituting the point (6, 1) into the circle's equation:

The circle passes through the point (6, 1), so we substitute \((x = 6, y = 1)\) into the equation:

\(6^2 + 1^2 - 2g(6) + 6(1) - 19c = 0\)

Simplifying gives:

\(36 + 1 - 12g + 6 - 19c = 0\)

\(43 - 12g - 19c = 0\)

Therefore, \(12g + 19c = 43\).

1. Using the condition for the center on the line:

The center of the circle \((g, -3)\) lies on the line \(x - 2cy = 8\).

Substituting \((x = g, y = -3)\) gives:

\(g - 2c(-3) = 8\)

\(g + 6c = 8\)

We have two equations now:

\(12g + 19c = 43\)  ...(i)

\(g + 6c = 8\)  ...(ii)

Let's solve these simultaneous equations:

1. Solving the equations:

From equation (ii):

\(g = 8 - 6c\)

Substituting in equation (i) gives:

\(12(8 - 6c) + 19c = 43\)

\(96 - 72c + 19c = 43\)

\(96 - 53c = 43\)

\(53c = 53\)

\(c = 1\)

\(g = 8 - 6(1) = 2\)

1. Finding the radius and x-intercepts:

Plug \((g, c)\) into the radius formula:

\(r = \sqrt{g^2 + (-3)^2 + 19c} = \sqrt{2^2 + 9 + 19(1)}\)

\(r = \sqrt{4 + 9 + 19} = \sqrt{32}\)

Now the generalized form for circle's equation is:

\((x - 2)^2 + (y + 3)^2 = 32\)

The length of the intercept made by the circle on the x-axis is:

\(2r_x = 2 \times \sqrt{c^2 - g^2} = 2 \times \sqrt{32 - 9} = 2 \times \sqrt{23}\)

Therefore, the length of intercept made by the circle on x-axis is \(2\sqrt{23}\).