Equation of circle which touches the line x = 0, y = 0 and x = 4?
(x-2)² + (y±2)² = 4
Let the centre of the required circle be (h, k) with radius r.
Since the circle touches the line x = 0, the perpendicular distance from the centre to this line equals r: \(|h| = r\).
Since the circle touches the line y = 0, similarly \(|k| = r\).
Since the circle touches the line x = 4, the distance from the centre to this line equals r: \(|h-4| = r\).
Taking h and r positive and the centre lying between the lines x = 0 and x = 4, h = r gives \(|r-4| = r \Rightarrow 4-r = r \Rightarrow r = 2\), so h = 2.
From \(|k| = r = 2\), k can be either +2 or −2, since nothing in the given conditions fixes whether the circle lies above or below the x-axis.
Hence two circles satisfy all three tangency conditions: \((x-2)^2+(y-2)^2=4\) and \((x-2)^2+(y+2)^2=4\), which together are written compactly as \((x-2)^2+(y\pm2)^2=4\).
Important Questions from Equation of Circle
The equation x2 + y2 + 2gx + 2fy + c = 0 always represents a circle whose centre is (-g, -f) and radius is \(\sqrt {{g^2} + {f^2} - c} \). If g2 + f2 = c, then in this case, the circle is called as:
Find the equation of the circle which passes through (-1, 1) and (2, 1), and having centre on the line x + 2y + 3 = 0.
A circle is drawn on the chord of a circle x 2+ y 2= a 2as diameter. The chord lies on the line x + y = a. What is the equation of the circle?
- If the points lie on a circle, then what is/are the possible value(s) of k ?
- If a circle is drawn through A, B and D, then what is the diameter of the circle ?
