The Total solution for NCERT class 6-12
At what point onthe circle x2 + y2 – 2x – 4y + 1 = 0, thetangent is parallel to x – axis.
⇒ –(x – 1) = 0
⇒ x= 1
Substituting x = 1 in x2 + y2 –2x – 4y + 1 = 0, we get,
⇒ 12 +y2 – 2(1) – 4y + 1 = 0
⇒ 1– y2 – 2 – 4y + 1 = 0
⇒ y2 –4y = 0
⇒ y(y – 4) = 0
⇒ y= 0 and y = 4
Thus, the required point is (1, 0) and (1, 4)