Problem 114 - A Complete Resource Book in Mathematics for JEE Main

Question

If a circle passes through the point \((a, b)\) and cuts the circle \(x^{2}+y^{2}=4\) orthogonally, then the locus of its centre is (A) \(2 a x+2 b y+\left(a^{2}+b^{2}+4\right)=0\) (B) \(2 a x+2 b y-\left(a^{2}+b^{2}+4\right)=0\) (C) \(2 a x-2 b y+\left(a^{2}+b^{2}+4\right)=0\) (D) \(2 a x-2 b y-\left(a^{2}+b^{2}+4\right)=0\)

Step-by-Step Solution

Answer

The locus of the center is given by option (A): \(2ax + 2by + (a^2 + b^2 + 4) = 0\).

Step 1: Understand the problem

We need to find the locus of the centre of a circle that passes through the point \((a, b)\) and cuts another circle \(x^2 + y^2 = 4\) orthogonally.

Step 2: Circle equation form

Consider the arbitrary circle with equation \((x-h)^2 + (y-k)^2 = r^2\) where \((h, k)\) is the center. The given circle equation is \(x^2 + y^2 = 4\), which can be rewritten as \((x-0)^2 + (y-0)^2 = 2^2\).

Step 3: Orthogonality condition

For two circles to cut orthogonally, the sum of the square of the radii must be equal to the square of the distance between their centers: \(d^2 = r_1^2 + r_2^2\). Here, \(d^2 = h^2 + k^2\), \(r_1^2 = 4\) and \(r_2^2 = r^2\). Thus, \(h^2 + k^2 = 4 + r^2\).

Step 4: Pass through point condition

The circle passes through the point \((a, b)\), so substitute \((a, b)\) into the circle equation: \((a-h)^2 + (b-k)^2 = r^2\).

Step 5: Deriving locus equation

From step 3, we know \(r^2 = h^2 + k^2 - 4\). Substitute \(r^2\) in step 4's equation: \((a-h)^2 + (b-k)^2 = h^2 + k^2 - 4\). By expanding and rearranging, we obtain: \(h^2 - 2ah + a^2 + k^2 - 2bk + b^2 = h^2 + k^2 - 4\), \[-2ah - 2bk = -a^2 - b^2 + 4\].

Step 6: Obtain locus equation

Simplify the equation from step 5: \[2ah + 2bk = a^2 + b^2 + 4\]. Thus, the locus of \((h, k)\) is described by the equation: \[2ax + 2by = a^2 + b^2 + 4\].

Step 7: Identifying the correct answer

Compare the derived equation with the given options. The correct choice matches option (A): \[2ax + 2by + (a^2 + b^2 + 4) = 0\], emphasizing the rearrangement made to match the given structure.

Key Concepts

Orthogonal Circles

Orthogonal circles are an interesting phenomenon in geometry. Two circles are said to be orthogonal if they intersect at right angles. This means that at every intersection point, the tangents to each circle are perpendicular. To find if two circles are orthogonal, a special condition must be satisfied: the sum of the squares of their radii should be equal to the square of the distance between their centers. In mathematical terms, if we have two circles with centers at \(h_1, k_1\) and \(h_2, k_2\), and radii \(r_1\) and \(r_2\) respectively, then they are orthogonal if: \[ (h_1 - h_2)^2 + (k_1 - k_2)^2 = r_1^2 + r_2^2 \].
In the given exercise, the arbitrary circle is compared with a fixed circle with the equation \(x^2 + y^2 = 4\) to apply this orthogonality condition. The center of the fixed circle is \( (0, 0) \), and its radius is \(2\). This forms the basis for further derivations.

Circle Equations

The circle equation is a foundational concept in geometry, representing the set of all points equidistant from a fixed point called the center. The general formula for a circle centered at \(h, k\) with radius \(r\) is given by: \[ (x-h)^2 + (y-k)^2 = r^2 \].
In the problem at hand, the given circle has the equation \(x^2 + y^2 = 4\). This can be rewritten in the standard form as \( (x-0)^2 + (y-0)^2 = 2^2 \). Here, the center is at the origin \( (0, 0) \) and the radius is \(2\).
Considering the arbitrary circle equation \( (x-h)^2 + (y-k)^2 = r^2 \), it must pass through the point \( (a, b) \). Using this condition, we substitute these values into the circle equation, helping us derive the equation that describes how the parameters are related.

Locus Derivation

The locus of a set of points is essentially a collection of points that satisfy a particular condition. For the given exercise, we are tasked to find the locus of the center \(h, k\) of a circle, passing through a specific point \( (a, b) \) and orthogonal to another specific circle.
The derivation starts with substituting the point \( (a, b) \) into the arbitrary circle equation, giving us \( (a-h)^2 + (b-k)^2 = r^2 \).
Coupled with the orthogonality condition \((h^2 + k^2 = 4 + r^2)\), we substitute for \(r^2\) and simplify the equation: \(-2ah - 2bk = -a^2 - b^2 + 4\). This is later rearranged to isolate the terms involving \(h\) and \(k\), giving us \[ 2ax + 2by = a^2 + b^2 + 4 \].
This final equation describes the locus, or the trajectory of possible positions, of the center of all circles satisfying the given conditions. The result is insightful as it ties together various geometric principles to illuminate the behavior of circle centers.