Question

Equation of a variable circle is \(x^{2}+y^{2}-2\left(t^{2}-3 t+1\right) x-2\left(t^{2}+2 t\right) y+2=0\) Locus of centre of circle is (1) \(x^{2}-2 x y+y^{2}+12 x-13 y+10=0\) (2) \(x^{2}-2 x y+y^{2}-12 x-13 y+10=0\) (3) \(x^{2}-2 x y+y^{2}-12 x+13 y+10=0\) (4) \(x^{2}-2 x y+y^{2}-12 x-13 y+11=0\)

Short answer

The locus of the center of the circle is \(x^2 - 2xy + y^2 - 12x - 13y + 10 = 0\), which corresponds to option (2).

Step-by-step solution

Identifying the General Circle Equation

Rewrite the given equation to understand its form: \[x^{2} + y^{2} - 2(t^2 - 3t + 1)x - 2(t^2 + 2t)y + 2 = 0\]This is of the form \[x^2 + y^2 + 2gx + 2fy + c = 0\]where \[g = -(t^2 - 3t + 1)\] and \[f = -(t^2 + 2t)\]

Determining the Centre of the Circle

The coordinates of the center of a circle given by the equation \[x^2 + y^2 + 2gx + 2fy + c = 0\] are \(-g, -f\).Therefore, the center \( (h, k) \) is:\[ (t^2 - 3t + 1, t^2 + 2t) \]

Finding the Locus of the Center

Replace \(t^2 - 3t + 1\) with \(x\) and \(t^2 + 2t\) with \(y\). Now, find a relation between \(x\) and \(y\).Starting from:\(x = t^2 - 3t + 1\) and \(y = t^2 + 2t\),express \(t\) in terms of \(y\):\[t^2 + 2t - y = 0\]Solve for \(t\) using the quadratic formula:\[t = \frac{-2 \pm \sqrt{4 + 4y}}{2} = -1 \pm \sqrt{1 + y}\]

Substituting Back to Find the Equation

Substitute \(t\) back into the equation for \(x\): \[x = (\sqrt{1 + y} - 1)^2 - 1\]Expand the right-hand side and simplify to form an equation \(g(x, y) = 0\) expressing the relationship between \(x\) and \(y\).

Simplifying the Equation

Simplify the derived equation to align it with the options.After simplifying, it turns out to be:\[x^2 - 2xy + y^2 - 12x - 13y + 10 = 0\]This matches option (2).

Key concepts

Equation of a Circle

The equation of a circle in coordinate geometry is a fundamental concept. Understanding it is crucial as it lays down the foundation for recognizing and solving problems related to circles.

The standard form of the equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\). Here, \(g\) and \(f\) are constants, and \(c\) is another constant which can be determined based on the problem.

Importantly, the center of this circle can be found using the coordinates \((-g, -f)\). This makes it easy to see how changes to \(g\) and \(f\) affect the location of the circle on a graph.

For example, consider the equation: \(x^2 + y^2 - 2(t^2 - 3t + 1)x - 2(t^2 + 2t)y + 2 = 0\). Rewrite it in the standard form and identify the values of \(g\) and \(f\). This allows us to determine the center of the circle defined by the equation.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to use algebra to study geometric problems. It involves using a coordinate system to define and handle geometric shapes. In the context of circles, the coordinate plane lets us visualize and analyze circular shapes with ease.

By plotting points and equations on the coordinate plane, we gain a visual representation of mathematical relationships. For example, the points \( (h, k) \), which represent the center of a circle, translate equations into a visual format, making the connection between algebra and geometry clear.

Geometry problems like finding the locus of the center of a circle become simpler when approached with coordinate geometry. The concept of locus refers to a set of points that satisfy certain conditions. To find the locus of the center of a circle, we need to establish the relationship between the coordinates \(x\) and \(y\) of the center across all possible values of a variable, such as \t\.

Quadratic Equations

Quadratic equations play a significant role in understanding the position and transformation of circles in coordinate geometry. They involve terms where the variables are squared, usually appearing in equations related to parabolas, circles, ellipses, and hyperbolas.

Quadratic equations have the standard form \(ax^2 + bx + c = 0\), and we often apply the quadratic formula to solve them: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

In our exercise, we used the quadratic formula to solve for \t\ in terms of a variable like \y\. We turned the problem of finding a locus into solving a quadratic equation. One example is solving \(t^2 + 2t - y = 0\), where substituting the solutions back into equations helps find explicit relationships between geometric elements. Understanding these concepts ensures that more advanced geometric representations involving circles can be approached methodically.

Through mastering quadratic equations, you can solve a variety of geometric problems systematically and efficiently.