What is the equation of the circle touching the lines x-3y-11=0 and 3x-y-9=0 having its center on the line x+2y+19=0
To find the equation of the circle that touches the given lines and has its center on the given line, we can use the concept of the distance between a point and a line.
Step 1: Find the intersection point of the two given lines.
To find the intersection point, we need to solve the system of equations formed by the two given lines:
x - 3y - 11 = 0 ...(Equation 1)
3x - y - 9 = 0 ...(Equation 2)
By solving these equations simultaneously, we can find the intersection point, which is (x₁, y₁).
Step 2: Find the equation of the line passing through the intersection point with the given slope.
The given line is: x + 2y + 19 = 0
The slope-intercept form of a line is y = mx + b, where m is the slope of the line and b is the y-intercept. To find the equation of the line passing through the intersection point, we need to substitute the values of (x₁, y₁) into the equation of the given line and solve for m and b.
Step 3: Find the perpendicular bisector of the line segment joining the intersection point and the center of the circle.
The perpendicular bisector of a line segment is a line that passes through the midpoint of the line segment and is perpendicular to it. To find the perpendicular bisector, we need to find the midpoint and slope.
Step 4: Find the radius of the circle.
The radius of the circle is the distance between the center and any point on the circle. Since the circle touches the given lines, the distance between the center and either line should be equal to the radius.
Step 5: Write the equation of the circle.
Using the formula of the equation of a circle in the general form, (x - h)² + (y - k)² = r², we can substitute the values of the center coordinates (h, k) and the radius r to obtain the equation of the circle.
By following these steps, we can find the equation of the circle touching the given lines and having its center on the given line.