A circle touches each of the lines \(x - y = 0\) and \(x + y = 0\) at unit distance from the origin. The centre of the circle may be at
\((\sqrt{2},\ 0)\)
Since \(x-y=0\) and \(x+y=0\) are mutually perpendicular lines through the origin, the centre of a circle touching both must lie on one of their angle bisectors, i.e. on the \(x\)-axis or \(y\)-axis. Taking the centre as \((a,0)\), the foot of the perpendicular to \(x-y=0\) is \(\left(\dfrac{a}{2},\dfrac{a}{2}\right)\), at distance \(\dfrac{a}{\sqrt{2}}\) from the origin. Since this point of tangency is at unit distance from the origin, \(\dfrac{a}{\sqrt{2}}=1\), giving \(a=\sqrt{2}\), so the centre may be at \((\sqrt{2},0)\).
Similar Questions
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?
The circle x 2+ y 2+ 4x – 7y + 12 = 0 cut an intercept on y- axis equal
What is the equation of the circle which passes through the points (3, -2) and (-2, 0) and having its centre on the line 2x – y – 3 = 0?
- 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 ?
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?
The circle x 2+ y 2+ 4x – 7y + 12 = 0 cut an intercept on y- axis equal
If the equation x 2+ y 2 - 4x - 4y + 4 = 0 represents a circle, then its radius is
