Point of intersection of two given lines is (1,1). Since each of the two given lines contains a diameter of the given circle, therefore the point of intersection of the two given lines is the centre of the given circle. Hence centre =(1,1) ∴a2−7a+11=1⇒a=2,5 and a2−6a+6=1⇒a=1,5 From both (i) and (ii), a=5 Now on replacing each of (a2−7a+11) and (a2−6a+6) by 1 , the equation of the given circle is x2+y2−2x−2y+b3+1=0 ⇒(x−1)2+(y−1)2+b3=1⇒b3=1−[(x−1)2+(y−1)2]∴b∈(−∞,1)
If each of the lines 5 x+8 y=13 and 4 x-y=3 contains a diameter of the circle x^2+y^2-2 (a^2-7 a+11 ) x-2 (a^2-6 a+6 ) y+b^3+1=0, then :
JEE Main 2013 — Mathematics Coordinate Geometry
If each of the lines 5x+8y=13 and 4x−y=3 contains a diameter of the circle x2+y2−2(a2−7a+11) x−2(a2−6a+6)y+b3+1=0, then :
Held on 9 Apr 2013 · Verified 6 Jul 2026.
Options
- A
a=5 and b∈/(−1,1)
- B
a=1 and b∈/(−1,1)
- C
a=2 and b∈/(−∞,1)
- D
a=5 and b∈(−∞,1)
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