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Exercise 12.2 · Q2

Q.Find the equation of the circle drawn on a diagonal of the rectangle as its diameter whose sides are the lines x=4x = 4, x=−5x = -5, y=5y = 5 and y=−1y = -1.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★est
30% · 12/40 Questions
✓ Free question

Find the rectangle's vertices from the four given lines, identify a diagonal, and use the diameter-form circle equation.

Circle on the segment joining (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) as diameter: (x−x1)(x−x2)+(y−y1)(y−y2)=0(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.

  1. The rectangle's sides are x=4x=4, x=−5x=-5, y=5y=5, y=−1y=-1, so its vertices are (4,5), (4,−1), (−5,5), (−5,−1)(4,5),\ (4,-1),\ (-5,5),\ (-5,-1).
  2. Take one diagonal, e.g. from (4,5)(4,5) to (−5,−1)(-5,-1) (opposite corners).
  3. Apply the diameter form with (x1,y1)=(4,5)(x_1,y_1)=(4,5) and (x2,y2)=(−5,−1)(x_2,y_2)=(-5,-1):

(x−4)(x−(−5))+(y−5)(y−(−1))=0(x-4)(x-(-5))+(y-5)(y-(-1))=0

  1. Expand: (x−4)(x+5)=x2+5x−4x−20=x2+x−20(x-4)(x+5)=x^2+5x-4x-20=x^2+x-20; (y−5)(y+1)=y2+y−5y−5=y2−4y−5(y-5)(y+1)=y^2+y-5y-5=y^2-4y-5.
  2. Add: x2+x−20+y2−4y−5=0⇒x2+y2+x−4y−25=0x^2+x-20+y^2-4y-5=0\Rightarrow x^2+y^2+x-4y-25=0.
  3. Self-check: midpoint of (4,5)(4,5) and (−5,−1)(-5,-1) is (−12,2)\left(-\tfrac12,2\right); from the equation, centre =(−12,2)=(-\tfrac12,2) ✓ (since 2g=1⇒g=122g=1\Rightarrow g=\tfrac12, centre x=−g=−12x=-g=-\tfrac12; 2f=−4⇒f=−22f=-4\Rightarrow f=-2, centre y=−f=2y=-f=2).
✓Final answer

Equation of the circle: x2+y2+x−4y−25=0x^2+y^2+x-4y-25=0.

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