Question

Consider $$ \begin{aligned} &L_{1}: 2 x+3 y+p-3=0 \\ &L_{2}: 2 x+3 y+p+3=0 \end{aligned} $$ where \(p\) is a real number, and \(C: x^{2}+y^{2}+6 x-10 y+30=0\). STATEMENT-1: If line \(L_{1}\) is a chord of circle \(C\), then line \(L_{2}\) is not always a diameter of circle \(C\). and STATEMENT-2: If line \(L_{1}\) is a diameter of circle \(C\), then line \(L_{2}\) is not a chord of circle \(C\). (A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True

Short answer

STATEMENT-1 is True, STATEMENT-2 is False

Step-by-step solution

Analyze the Circle's Equation

First, rewrite the equation of the circle in the standard form by completing the square for both x and y. The given equation is: \(x^2 + 6x + y^2 - 10y + 30 = 0\). Completing the square gives \((x+3)^2 - 9 + (y-5)^2 - 25 + 30 = 0\) which simplifies to \((x+3)^2 + (y-5)^2 = 4\). The center of the circle is \((-3, 5)\) and the radius is 2 units.

Evaluate Lines as Chords or Diameters

For a line to be a chord of a circle, it must intersect the circle at two distinct points. For it to be a diameter, it must pass through the center of the circle. Since lines \(L_1\) and \(L_2\) have the same slope, they are parallel and cannot both pass through the center to be diameters. To check if \(L_1\) can be a chord, substitute the center of the circle into the line's equation and verify if it satisfies the condition for intersecting the circle.

Check Statement-1

To check if line \(L_1\) is a chord, we substitute the center \((-3, 5)\) into the equation of \(L_1\). Substituting, we get \(2(-3)+3(5)+p-3=0\) which simplifies to \(p=6\). For this value of p, \(L_1\) goes through a point on the circle but does not necessarily pass through the center. Thus, \(L_1\) can be a chord of the circle, and \(L_2\) parallel to it, is not necessarily a diameter. Statement-1 is true.

Check Statement-2

For line \(L_1\) to be a diameter, it must pass through the center \((-3, 5)\). Substituting the center of the circle into the equation of \(L_1\) gives a condition for \(p\) that we found earlier: \(p=6\). Substituting this \(p\) into \(L_2\), we get \(2(-3)+3(5)+6+3=0\), which simplifies to \(0=0\), satisfying the condition. This means for \(p=6\), \(L_1\) and \(L_2\) are actually the same line, which contradicts the circle having a separate diameter and a chord with the given equations. Hence, Statement-2 is false.

Key concepts

Chord of a Circle

In geometry, a chord of a circle is a straight line segment whose endpoints both lie on the circle. It's worth noting that all diameters are chords, but not all chords are diameters.

Now, to determine whether a given line is a chord of a circle, the line must intersect the circle at precisely two distinct points. The provided exercise involves analyzing whether the line equation given as \(L_1: 2x + 3y + p - 3 = 0\) can qualify as a chord of the given circle \(C: x^2 + y^2 + 6x - 10y + 30 = 0\). By checking if this line intersects our circle at two points, which we can do by substituting the center of the circle into the line's equation and verifying if it satisfies the circle's equation, we can conclude whether \(L_1\) is a chord. It does not need to pass through the center of the circle, which is a key difference between a chord and a diameter.

To teach these concepts more effectively, we advise demonstrating this intersection by solving the system of equations graphically or algebraically, showing students the intersection points.

Diameter of a Circle

The diameter of a circle is the longest possible chord and it passes through the center of the circle. It is also exactly twice the length of the radius. So a necessary and sufficient condition for a line to be a diameter of a circle is that it must pass through the center of that circle.

In our exercise, we need to determine if \(L_2: 2x + 3y + p + 3 = 0\) can be a diameter of circle \(C\). Based on the definition of a diameter, we know it must go through the circle's center, which in this case, is \( (-3, 5) \). By substituting this center point into the equation of \(L_2\), we can ascertain if it satisfies this condition. If it does not, then we can conclusively state that \(L_2\) is not a diameter. Encouraging students to use this substitution method helps instill the understanding that the diameter's unique position is essential to its definition.

Equation of a Circle

Getting to grips with the equation of a circle is essential in circle geometry. The general form of the equation is \( (x - h)^2 + (y - k)^2 = r^2 \) where \( (h,k) \) is the center and \( r \) is the radius. To find the standard form from the expanded equation like \( x^2 + y^2 + 6x - 10y + 30 = 0 \), one technique used is 'completing the square' which can simplify the equation and thus make it easier to identify the circle's characteristics.

Students can learn better when shown step-by-step how to complete the square. For our exercise, this technique allows us to determine the circle's center \( (-3, 5) \) and radius 2, after which we can assess the nature of \(L_1\) and \(L_2\). It's invaluable to emphasize the power of transforming the equation into its standard form as it makes solving numerous circle geometry problems much more straightforward.