Question

If the lines $x + y = 6$ and $x + 2 y = 4$ be diameters of the circle whose diameter is 20, then the equation of the circle is.

• A. $x ^ { 2 } + y ^ { 2 } - 16 x + 4 y - 32 = 0$
• B. $x ^ { 2 } + y ^ { 2 } + 16 x + 4 y - 32 = 0$
• C. $x ^ { 2 } + y ^ { 2 } + 16 x + 4 y + 32 = 0$
• D. $x ^ { 2 } + y ^ { 2 } + 16 x - 4 y + 32 = 0$

Answer 1

Answer: (A)

$x ^ { 2 } + y ^ { 2 } - 16 x + 4 y - 32 = 0$

Answer 2

Here $r = 10$ (radius)

Centre will be the point of intersection of the diameters, i.e. (8, –2). Hence required equation is

$( x - 8 ) ^ { 2 } + ( y + 2 ) ^ { 2 } = 10 ^ { 2 } \Rightarrow x ^ { 2 } + y ^ { 2 } - 16 x + 4 y - 32 = 0$.