The general equation of a circle with center (h,k) and radius r is given by (x-h)² + (y-k)² = r². We need to find the values of h, k, and r to determine the equation of the circle.

Find the equation of the circle which passes through the points $(2,-2)$ , and $(3{,}4)$ and whose centre lies on the line $x+y=2$ .

Answer: The equation of the circle is $\left(x-\frac{7}{10}\right)^2 + \left(y-\frac{13}{10}\right)^2 = \frac{629}{50}$ .

Step-by-step solution

Step 1: General Equation of a Circle

The general equation of a circle with center $(h,k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$ . We need to find the values of $h$ , $k$ , and $r$ to determine the equation of the circle.

Step 2: Formulate Equations from Given Points

Since the circle passes through the points $(2,-2)$ and $(3{,}4)$ , these points must satisfy the circle's equation. Substituting these coordinates into the general equation gives us two equations involving $h$ , $k$ , and $r$ .

Step 3: Equate the two expressions for r squared

Since both expressions are equal to $r^2$ , we can set them equal to each other. This eliminates $r$ and gives us an equation solely in terms of $h$ and $k$ . Expanding and simplifying this equation will help us find a relationship between $h$ and $k$ .

Step 4: Simplify the equation for h and k

Expanding the squared terms and simplifying the equation from the previous step, we get a linear equation relating $h$ and $k$ . This equation, $2h + 12k = 17$ , represents the perpendicular bisector of the segment connecting the two given points.

Step 5: Use the condition that the center lies on the line

We are given that the center $(h,k)$ lies on the line $x+y=2$ . Substituting $h$ for $x$ and $k$ for $y$ gives us another linear equation: $h+k=2$ . Now we have a system of two linear equations with two variables, $h$ and $k$ .

Step 6: Solve the system of linear equations

We solve the system of equations: $2h + 12k = 17$ and $h + k = 2$ . From the second equation, we express $h$ as $2-k$ and substitute it into the first equation. This allows us to find $k = \frac{13}{10}$ , and then $h = \frac{7}{10}$ . So, the center of the circle is $(\frac{7}{10}, \frac{13}{10})$ .

Step 7: Calculate the radius and final equation

Now that we have the center $(h,k) = (\frac{7}{10}, \frac{13}{10})$ , we can find $r^2$ by substituting these values and one of the given points (e.g., $(2,-2)$ ) into the circle equation. After calculating $r^2 = \frac{629}{50}$ , we write the final equation of the circle.

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Find the equation of the circle which passes through the points (2,−2)(2,-2)(2,−2), and (3,4)(3{,}4)(3,4) and whose centre lies on the line x+y=2x+y=2x+y=2.