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Worked Examples · Example 4

Q.Find the equation of the circle which passes through the points (2,−2)(2, -2) and (3,4)(3, 4) and whose centre lies on the line x+y=2x + y = 2.

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The key idea is to use the standard circle equation (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2, substitute the given points, and use the centre condition h+k=2h+k=2 to solve for hh, kk, and rr. The required circle equation is (x−0.7)2+(y−1.3)2=12.58(x-0.7)^2+(y-1.3)^2=12.58.

We need the equation of a circle that passes through two specific points and has its centre on a given line. The most direct way is to use the standard form of a circle: (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2, where (h,k)(h,k) is the centre and rr is the radius. The problem gives us two conditions from the points, and a third condition from the line — three unknowns, three equations.

Let’s set it up step by step.

  1. Write the general equation and apply the first point The circle passes through (2,−2)(2,-2). Substituting into (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2:

(2−h)2+(−2−k)2=r2(2-h)^2+(-2-k)^2=r^2

This simplifies to:

(2−h)2+(k+2)2=r2(since (−2−k)2=(k+2)2)(2-h)^2+(k+2)^2=r^2 \quad \text{(since }(-2-k)^2=(k+2)^2\text{)}

  1. Apply the second point The circle also passes through (3,4)(3,4):

(3−h)2+(4−k)2=r2(3-h)^2+(4-k)^2=r^2

  1. Equate the two expressions for r2r^2 Since both equal r2r^2, we set them equal:

(2−h)2+(k+2)2=(3−h)2+(4−k)2(2-h)^2+(k+2)^2=(3-h)^2+(4-k)^2

Expand each square:

  • Left: (4−4h+h2)+(k2+4k+4)=h2+k2−4h+4k+8(4-4h+h^2)+(k^2+4k+4)=h^2+k^2-4h+4k+8
  • Right: (9−6h+h2)+(16−8k+k2)=h2+k2−6h−8k+25(9-6h+h^2)+(16-8k+k^2)=h^2+k^2-6h-8k+25

Cancel h2+k2h^2+k^2 from both sides:

−4h+4k+8=−6h−8k+25-4h+4k+8=-6h-8k+25

Bring terms together:

(−4h+6h)+(4k+8k)=25−8(-4h+6h)+(4k+8k)=25-8

2h+12k=172h+12k=17

So we have:

h+6k=172(Equation 1)h+6k=\frac{17}{2} \quad \text{(Equation 1)}

  1. Use the centre condition The centre (h,k)(h,k) lies on x+y=2x+y=2, so:

h+k=2(Equation 2)h+k=2 \quad \text{(Equation 2)}

  1. Solve for hh and kk Subtract Equation 2 from Equation 1:

(h+6k)−(h+k)=172−2(h+6k)-(h+k)=\frac{17}{2}-2

5k=132⇒k=1310=1.35k=\frac{13}{2} \quad \Rightarrow \quad k=\frac{13}{10}=1.3

Then from h+k=2h+k=2:

h=2−1.3=0.7h=2-1.3=0.7 …

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