Question

A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation $2 x - y = 2$ Then the equation of the circle is

• A. $x ^ { 2 } + y ^ { 2 } + 2 x - 1 = 0$
• B. $x ^ { 2 } + y ^ { 2 } - 2 x - 1 = 0$
• C. $x ^ { 2 } + y ^ { 2 } - 2 y - 1 = 0$
• D. None of these

Answer 1

Answer: (B)

$x ^ { 2 } + y ^ { 2 } - 2 x - 1 = 0$

Answer 2

The line joining (4, 3) and (2, 1) is also along a diameter. So, the centre is the intersection of the diameters $2 x - y = 2$ and $y - 3 = ( x - 4 )$ . Solving these, the centre = (1, 0)

$\therefore$ Radius = Distance between (1, 0) and (2, 1) =

$\therefore$Equation of circle $( x - 1 ) ^ { 2 } + y ^ { 2 } = ( \sqrt { 2 } ) ^ { 2 }$

$\Rightarrow x ^ { 2 } + y ^ { 2 } - 2 x - 1 = 0$