Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Circle question

2005 · Shift 0 · Q104
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Circle
  5. /2005 · Shift 0 · Q104

Circle question

2005 · Shift 0 · Q104

JEE MainMathematicsCircleMCQ+4 / −1
If a circle passes through the point (a, b) and cuts the circle x2 + y2=p2{x^2}\, + \,{y^2} = {p^2}x2+y2=p2 orthogonally, then the equation of the locus of its centre is :
  1. A
    x2 + y2− 3ax − 4 by  + (a2 + b2−p2)=0{x^2}\, + \,{y^2} - \,3ax\, - \,4\,by\,\, + \,({a^2}\, + \,{b^2} - {p^2}) = 0x2+y2−3ax−4by+(a2+b2−p2)=0
  2. B
    2ax +  2 by  − (a2 − b2+p2)=02ax\, + \,\,2\,by\,\, - \,({a^2}\, - \,{b^2} + {p^2}) = 02ax+2by−(a2−b2+p2)=0
  3. C
    x2 + y2− 2ax −  3 by  + (a2 − b2−p2)=0{x^2}\, + \,{y^2} - \,2ax\, - \,\,3\,by\,\, + \,({a^2}\, - \,{b^2} - {p^2}) = 0x2+y2−2ax−3by+(a2−b2−p2)=0
  4. D
    2ax +  2 by  − (a2 + b2+p2)=02ax\, + \,\,2\,by\,\, - \,({a^2}\, + \,{b^2} + {p^2}) = 02ax+2by−(a2+b2+p2)=0
View written solutionFree

Correct answer: D

  1. Let the centre of the required circle be (x,y).(x,y).(x,y). Let its radius be rrr.

  2. Since the circle passes through (a,b)(a,b)(a,b), its radius satisfies r2=(x−a)2+(y−b)2.r^2=(x-a)^2+(y-b)^2.r2=(x−a)2+(y−b)2.

  3. Condition for orthogonal intersection with x2+y2=p2x^2+y^2=p^2x2+y2=p2 whose centre is (0,0)(0,0)(0,0) and radius is ppp:

    If two circles with centres distance ddd and radii r1,r2r_1,r_2r1​,r2​ cut orthogonally, then d2=r12+r22.d^2=r_1^2+r_2^2.d2=r12​+r22​.

    Here, d2=x2+y2, r1=r, r2=p.d^2=x^2+y^2,\, r_1=r,\, r_2=p.d2=x2+y2,r1​=r,r2​=p. So, x2+y2=r2+p2.x^2+y^2=r^2+p^2.x2+y2=r2+p2.

  4. Substitute r2=(x−a)2+(y−b)2r^2=(x-a)^2+(y-b)^2r2=(x−a)2+(y−b)2: x2+y2=(x−a)2+(y−b)2+p2.x^2+y^2=(x-a)^2+(y-b)^2+p^2.x2+y2=(x−a)2+(y−b)2+p2.

  5. Expand the right-hand side: (x−a)2+(y−b)2=x2−2ax+a2+y2−2by+b2.(x-a)^2+(y-b)^2=x^2-2ax+a^2+y^2-2by+b^2.(x−a)2+(y−b)2=x2−2ax+a2+y2−2by+b2.

    Therefore, x2+y2=x2+y2−2ax−2by+a2+b2+p2.x^2+y^2=x^2+y^2-2ax-2by+a^2+b^2+p^2.x2+y2=x2+y2−2ax−2by+a2+b2+p2.

  6. Cancel x2+y2x^2+y^2x2+y2 from both sides: 0=−2ax−2by+a2+b2+p2.0=-2ax-2by+a^2+b^2+p^2.0=−2ax−2by+a2+b2+p2.

    Rearranging, 2ax+2by−(a2+b2+p2)=0.2ax+2by-(a^2+b^2+p^2)=0.2ax+2by−(a2+b2+p2)=0.

  7. Hence, the locus of the centre is 2ax+2by−(a2+b2+p2)=0.\boxed{2ax+2by-(a^2+b^2+p^2)=0}. 2ax+2by−(a2+b2+p2)=0​.

  8. Compare with options: This matches Option D.

  9. Comparison with stored correct answer: Stored correct answer is D, which agrees with the derived result.

PreviousNext

More from Circle

  • If the circles x2+y2+2ax+cy+a=0 and x2+y2−3ax+dy−1=0 intersect in two ditinct points P and Q then the line 5x + by - a = 0 passes through P and Q for :2005 · MCQ
  • If the pair of lines ax2+2(a+b)xy+by2=0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then :2005 · MCQ
  • If the lines 2x + 3y + 1 + 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference 10π, then the equation of the circle is :2004 · MCQ
  • If a circle passes through the point (a, b) and cuts the circle x2+y2=4 orthogonally, then the locus of its centre is :2004 · MCQ
  • Intercept on the line y = x by the circle x2+y2−2x=0 is AB. Equation of the circle on AB as a diameter is :2004 · MCQ
  • The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is :2003 · MCQ
  • If the chord y = mx + 1 of the circle x2+y2=1 subtends an angle of measure 45∘ at the major segment of the circle then value of m is :2002 · MCQ
  • The equation of a circle with origin as a center and passing through an equilateral triangle whose median is of length 3a is :2002 · MCQ