The circle x2+y2 - 8x + 4y + 4 = 0 touches

Question

The circle x2+y2 - 8x + 4y + 4 = 0 touches (A) x - axis (B) y - axis (C) both axis (D) neither x-axis nor y-axis

Tardigrade Admin · Accepted Answer

beginarrayl x2+y2-8 x+4 y+4=0 ⇒(x2-8 x+16)+(y2+4 y+4)=16 ⇒(x-4)2+(y+2)2=42 endarray Center is (4,-2) and radius is 4 Hence given circle touches y-axis

Q. The circle $x^{2} + y^{2} - 8 x + 4 y + 4 = 0$ touches

x - axis

y - axis

both axis

neither x-axis nor y-axis

$< b r / > < b r / > x^{2} + y^{2} - 8 x + 4 y + 4 = 0 < b r / >\Rightarrow (x^{2} - 8 x + 16) + (y^{2} + 4 y + 4) = 16 < b r / >\Rightarrow (x - 4)^{2} + (y + 2)^{2} = 4^{2} < b r / > < b r / >$
Center is $(4, - 2)$ and radius is 4
Hence given circle touches $y$ -axis

Q. The circle x2+y2−8x+4y+4=0 touches