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Circle question

2013 · Shift 2 · Q34
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Circle question

2013 · Shift 2 · Q34

JEE AdvancedMathematicsCircleMultiple correct+4 / −1
Circle (s) touching x-axis at a distance 3 from the origin and having an intercept of length 272\sqrt 727​ on y-axis is (are)
  1. A
    x2 + y2 − 6x  +8y +9=0{x^2}\, + \,{y^2}\, - \,6x\,\, + 8y\, + 9 = 0x2+y2−6x+8y+9=0
  2. B
    x2 + y2 − 6x  +7y +9=0{x^2}\, + \,{y^2}\, - \,6x\,\, + 7y\, + 9 = 0x2+y2−6x+7y+9=0
  3. C
    x2 + y2 − 6x  −8y +9=0{x^2}\, + \,{y^2}\, - \,6x\,\, - 8y\, + 9 = 0x2+y2−6x−8y+9=0
  4. D
    x2 + y2 − 6x  −7y +9=0{x^2}\, + \,{y^2}\, - \,6x\,\,- 7y\, + 9 = 0x2+y2−6x−7y+9=0
View written solutionFree

Correct answer: A, C

  1. Interpret the condition “touching x-axis at a distance 3 from the origin”

If a circle touches the xxx-axis, then its point of contact lies on the xxx-axis.

Given that this point is at a distance 333 from the origin, the point of contact must be either

(3,0)or(−3,0).(3,0) \quad \text{or} \quad (-3,0).(3,0)or(−3,0).

Now look at the options: all have equation of the form

x2+y2−6x+⋯=0,x^2+y^2-6x+\cdots=0,x2+y2−6x+⋯=0,

which gives center with xxx-coordinate 333. So the relevant point of contact is (3,0)(3,0)(3,0).

Hence the center must be of the form

(3,k)(3,k)(3,k)

and radius must be

r=∣k∣r=|k|r=∣k∣

because the circle touches the xxx-axis.


  1. Use the intercept condition on the y-axis

The length of intercept cut by the circle on the yyy-axis is 272\sqrt{7}27​.

For a circle with center (3,k)(3,k)(3,k) and radius r=∣k∣r=|k|r=∣k∣, the distance of the center from the yyy-axis is 333.

So the chord cut by the line x=0x=0x=0 has length

2r2−32=2r2−9.2\sqrt{r^2-3^2}=2\sqrt{r^2-9}.2r2−32​=2r2−9​.

Given this equals 272\sqrt{7}27​,

2r2−9=272\sqrt{r^2-9}=2\sqrt{7}2r2−9​=27​

which gives

r2−9=7r^2-9=7r2−9=7 r2=16r^2=16r2=16 r=4.r=4.r=4.

Thus

∣k∣=4⇒k=4 or −4.|k|=4 \Rightarrow k=4 \text{ or } -4.∣k∣=4⇒k=4 or −4.

So the possible circles have centers

(3,4)or(3,−4),(3,4) \quad \text{or} \quad (3,-4),(3,4)or(3,−4),

with radius 444.


  1. Write equations of the possible circles

Case 1: Center (3,4)(3,4)(3,4), radius 444

(x−3)2+(y−4)2=16(x-3)^2+(y-4)^2=16(x−3)2+(y−4)2=16

Expanding:

x2−6x+9+y2−8y+16=16x^2-6x+9+y^2-8y+16=16x2−6x+9+y2−8y+16=16 x2+y2−6x−8y+9=0.x^2+y^2-6x-8y+9=0.x2+y2−6x−8y+9=0.

This matches Option C.

Case 2: Center (3,−4)(3,-4)(3,−4), radius 444

(x−3)2+(y+4)2=16(x-3)^2+(y+4)^2=16(x−3)2+(y+4)2=16

Expanding:

x2−6x+9+y2+8y+16=16x^2-6x+9+y^2+8y+16=16x2−6x+9+y2+8y+16=16 x2+y2−6x+8y+9=0.x^2+y^2-6x+8y+9=0.x2+y2−6x+8y+9=0.

This matches Option A.


  1. Check the remaining options

Option B:

x2+y2−6x+7y+9=0x^2+y^2-6x+7y+9=0x2+y2−6x+7y+9=0

Compare with

x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0

Here,

2g=−6⇒g=−3,2f=7⇒f=72.2g=-6 \Rightarrow g=-3, \qquad 2f=7 \Rightarrow f=\frac{7}{2}.2g=−6⇒g=−3,2f=7⇒f=27​.

Center is

(−g,−f)=(3,−72)(-g,-f)=\left(3,-\frac{7}{2}\right)(−g,−f)=(3,−27​)

and radius

r=g2+f2−c=9+494−9=72.r=\sqrt{g^2+f^2-c}= \sqrt{9+\frac{49}{4}-9}=\frac{7}{2}.r=g2+f2−c​=9+449​−9​=27​.

This does touch the xxx-axis, but the intercept on the yyy-axis is

2r2−32=2494−9=2134=13,2\sqrt{r^2-3^2}=2\sqrt{\frac{49}{4}-9}=2\sqrt{\frac{13}{4}}=\sqrt{13},2r2−32​=2449​−9​=2413​​=13​,

which is not 272\sqrt{7}27​. So B is incorrect.

Option D:

x2+y2−6x−7y+9=0x^2+y^2-6x-7y+9=0x2+y2−6x−7y+9=0

Center is

(3,72),\left(3,\frac{7}{2}\right),(3,27​),

radius is again 72\frac{7}{2}27​. Its intercept on the yyy-axis is also

13,\sqrt{13},13​,

not 272\sqrt{7}27​. So D is incorrect.


  1. Final answer

The correct circles are:

A, C\boxed{A,\ C}A, C​
  1. Comparison with stored correct answer

Stored correct answer: A, C

My derived answer: A, C

So, the answer agrees with the stored correct answer.

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