If the lines $3 x-4 y-7=0$ and $2 x-3 y-5=0$ are two diameters of a circle of area $49 \pi$ square units, the equation of the circle is

(D) $x^{2}+y^{2}-2 x+2 y-47=0$

(A) $x^{2}+y^{2}+2 x-2 y-47=0$

(B) $x^{2}+y^{2}+2 x-2 y-62=0$

(C) $x^{2}+y^{2}-2 x+2 y-62=0$

JEE Main · 2006 · 30 Apr 2006 · Offline · Circle

If the lines

3 x - 4 y - 7 = 0

and

2 x - 3 y - 5 = 0

are two diameters of a circle of area

49 π

square units, the equation of the circle is

(A) $x^{2} + y^{2} + 2 x - 2 y - 47 = 0$
(B) $x^{2} + y^{2} + 2 x - 2 y - 62 = 0$
(C) $x^{2} + y^{2} - 2 x + 2 y - 62 = 0$
(D) $x^{2} + y^{2} - 2 x + 2 y - 47 = 0$ ✓ Correct

Answer: (D)

Solution

The two given diameters meet at the centre of the circle, so solve

3 x - 4 y - 7 = 0

and

2 x - 3 y - 5 = 0

together. Taking twice the first minus three times the second eliminates

x

and gives

y = - 1

; substituting back gives

3 x + 4 = 7

, so

x = 1

, and the centre is

(1, - 1)

. The area

49 π

means

π r^{2} = 49 π

, so

r = 7

. The circle is

(x - 1)^{2} + (y + 1)^{2} = 49

, which expands to

x^{2} + y^{2} - 2 x + 2 y - 47 = 0

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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