Problem 103 - A Complete Resource Book in Mathematics for JEE Main

Question

Assertion: The locus of the centres of circles passing through the origin and cutting the circle \(x^{2}+y^{2}+6 x-\) \(4 y+2=0\) orthogonally is \(3 x-2 y+1=0\). Reason: The two circles \(x^{2}+y^{2}+2 g_{1} x+2 f_{1} y+c_{1}=0\) and \(x^{2}+y^{2}+2 g_{2} x+2 f_{2} y+c_{2}=0\) cut each other orthogonally if \(2 g_{1} g_{2}+2 f f_{2}=c_{1}+c_{2}\)

Step-by-Step Solution

Answer

The locus is correct: equation \(3x - 2y + 1 = 0\) is the answer.

Step 1: Analyze Given Circle Equation

The given circle equation is \(x^2 + y^2 + 6x - 4y + 2 = 0\). We need to identify the center \((-g_2, -f_2)\) and the equation constants \(g_2, f_2, c_2\). Comparing with the standard equation \(x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\), we find \(g_2 = 3\), \(f_2 = -2\), and \(c_2 = 2\).

Step 2: Setup Conditions for Orthogonality and Passing Through Origin

For two circles to cut orthogonally, \(2g_1g_2 + 2f_1f_2 = c_1 + c_2\). For the circle to pass through the origin: \(c_1 = 0\). Substitute \(c_1 = 0\) into the ortogonality condition, we get: \(2g_1(3) + 2f_1(-2) = 0 + 2\).

Step 3: Simplify Orthogonality Condition

Simplifying gives \(6g_1 - 4f_1 = 2\). Dividing through by 2 gives \(3g_1 - 2f_1 = 1\). This equation represents the condition on \((g_1, f_1)\), which are the coordinates of the centers of the circles (\(-g_1, -f_1\)) that cut the given circle orthogonally and pass through the origin.

Step 4: Interpret Results as Locus Equation

The equation \(3g_1 - 2f_1 = 1\) can be rewritten in terms of \(g_1\) and \(f_1\) as \(3(-x) - 2(-y) = 1\), which gives the locus \(3x - 2y + 1 = 0\). This is the required locus of the centers of the circles.

Key Concepts

Orthogonal Circles

Orthogonal circles are two circles that intersect at right angles, meaning their tangents at the points of intersection are perpendicular to each other. This is an interesting property because when two circles intersect orthogonally, certain geometric and algebraic conditions are satisfied.
One key algebraic condition for two circles to be orthogonal is that the sum of the products of their corresponding coefficients equals the sum of their constant terms.

• For circles with equations \(x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0\) and \(x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\), the orthogonality condition is: \(2g_1g_2 + 2f_1f_2 = c_1 + c_2\).
• This equation helps in determining if the two circles will intersect orthogonally without actually solving for their intersection points.

When you encounter problems involving orthogonal circles, verifying this condition can save time and provide insight into the relationship between the circles' geometry.

Circle Equations

Equations of circles are a central topic in coordinate geometry. The standard form of a circle's equation is \(x^2 + y^2 + 2gx + 2fy + c = 0\), where \((-g, -f)\) represent the center's coordinates and the radius can be derived using these constants.
The center-radius form \((x - h)^2 + (y - k)^2 = r^2\) is another popular format, where \((h, k)\) is the circle's center and \(r\) its radius.

• The given equation \(x^2 + y^2 + 6x - 4y + 2 = 0\) can be rearranged to match the standard form, identifying its center at \((-3, 2)\).
• Such transformations from general to standard form often involve completing the square to find the explicit center and radius.

Understanding circle equations allows us to easily navigate between these forms and interpret geometric properties like center and radius directly from the equation.

Circles Passing Through a Point

When dealing with circles passing through a specific point, such as the origin \((0, 0)\), certain aspects of the circle equation become constrained. In general, if a circle passes through a particular point, substituting the coordinates of that point into the circle's equation should satisfy the equation, often resulting in a condition on the circle's coefficients.
For example:

• If the circle is defined by \(x^2 + y^2 + 2gx + 2fy + c = 0\) and passes through the origin, substituting \((0, 0)\) gives \(c = 0\).
• This condition simplifies the determination of other unknowns impacting the circle's orientation and size.

In context, working with circles passing through the origin, especially when considering loci, provides a direct method to handle specific constraints, enabling a clearer path to derive solutions.