91Ó°ÊÓ

In geometry, understanding the equation of a circle is fundamental. A circle's equation is derived from the definition of a circle: a set of all points equidistant from a given point, known as the center. The standard form of the equation of a circle is given by: \[(x - h)^2 + (y - k)^2 = r^2\]where \((h, k)\) represents the center of the circle and \(r\) is the radius. The variables \(x\) and \(y\) are the coordinates of any point on the circle.
When working with problems involving circles, determining these parameters—center and radius—is key. In the given exercise, the circle passes through the point \((-2, 1)\) and is tangent to the line \(3x - 2y = 6\) at \((4, 3)\), allowing us to calculate these parameters step by step.
By using the relationship of the perpendicular line (discussed later), we find the circle's center at a calculated position \((6, 0)\). Once we have the circle's center and a tangent point, determining the radius becomes straightforward using the distance formula. The constructed equation is finally \((x - 6)^2 + y^2 = 37\), confirming the circle's shape and size.

A tangent line to a circle is a straight line that touches the circle at exactly one point without crossing it. This point is called the point of tangency. It's important to note that a tangent line is perpendicular to the radius drawn to the point of tangency—a key property in solving circles-related problems.
In our exercise, the tangent line is given by \(3x - 2y = 6\) and the point of tangency is \((4, 3)\). This information allows us to use the properties of perpendicular lines to find the equation of the line that passes through the center of the circle. This line is key to determining where the center will be placed on our circle plane. After correctly identifying the tangent line, the next step is finding its perpendicular counterpart, which further assists in locating the circle’s center.

Understanding perpendicular lines can be crucial when dealing with circles and their tangents. Two lines are perpendicular if the product of their slopes is \(-1\). For instance, if a line has a slope \(m\), a perpendicular line will have a slope of \(-1/m\).
In our exercise, the tangent line has a slope derived from its equation \(3x - 2y = 6\), rearranging to give a slope of \(\frac{3}{2}\). The perpendicular slope is therefore \(-\frac{3}{2}\), which helps us construct the perpendicular line passing through the point of tangency. This line's equation uses point-slope form \(y - y_1 = m(x - x_1)\) to find the circle center's path.
By understanding how perpendicular lines operate, solving geometry problems involving circles becomes much more manageable, as it directly relates to positioning the circle's center relative to its tangent point.