- Home
- Standard 11
- Mathematics
10-1.Circle and System of Circles
easy
The equation of circle with centre $(1, 2)$ and tangent $x + y - 5 = 0$ is
A
${x^2} + {y^2} + 2x - 4y + 6 = 0$
B
${x^2} + {y^2} - 2x - 4y + 3 = 0$
C
${x^2} + {y^2} - 2x + 4y + 8 = 0$
D
${x^2} + {y^2} - 2x - 4y + 8 = 0$
Solution

(b) Radius of circle = perpendicular distance of tangent from the centre of circle
$ \Rightarrow $ $r = \frac{{1 + 2 – 5}}{{\sqrt {1 + 1} }} = \sqrt 2 $
Hence the equation of required circle is
${(x – 1)^2} + {(y – 2)^2} = {(\sqrt 2 )^2}$
$ \Rightarrow \,{x^2} + {y^2} – 2x – 4y + 3 = 0.$
Standard 11
Mathematics