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Circle question

2024 · 9 Apr · Shift 1 · Q40
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  5. /2024 · 9 Apr · Shift 1 · Q40

Circle question

2024 · 9 Apr · Shift 1 · Q40

JEE MainMathematicsCircleMCQ+4 / −1
Let a circle passing through (2,0)(2,0)(2,0) have its centre at the point (h,k)(\mathrm{h}, \mathrm{k})(h,k). Let (xc,yc)(x_{\mathrm{c}}, y_{\mathrm{c}})(xc​,yc​) be the point of intersection of the lines 3x+5y=13 x+5 y=13x+5y=1 and (2+c)x+5c2y=1(2+\mathrm{c}) x+5 \mathrm{c}^2 y=1(2+c)x+5c2y=1. If h=lim⁡c→1xc\mathrm{h}=\lim_{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}h=c→1lim​xc​ and k=lim⁡c→1yc\mathrm{k}=\lim_{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}k=c→1lim​yc​, then the equation of the circle is :
  1. A
    5x2+5y2−4x−2y−12=05 x^2+5 y^2-4 x-2 y-12=05x2+5y2−4x−2y−12=0
  2. B
    25x2+25y2−20x+2y−60=025 x^2+25 y^2-20 x+2 y-60=025x2+25y2−20x+2y−60=0
  3. C
    25x2+25y2−2x+2y−60=025 x^2+25 y^2-2 x+2 y-60=025x2+25y2−2x+2y−60=0
  4. D
    5x2+5y2−4x+2y−12=05 x^2+5 y^2-4 x+2 y-12=05x2+5y2−4x+2y−12=0
View written solutionFree

Correct answer: B

  1. Find the intersection point  (xc,yc) \,(x_c,y_c)\,(xc​,yc​) of 3x+5y=1and(2+c)x+5c2y=1.3x+5y=1 \quad \text{and} \quad (2+c)x+5c^2y=1.3x+5y=1and(2+c)x+5c2y=1.

    Subtract the first equation from the second: [(2+c)−3]x+5(c2−1)y=0[(2+c)-3]x + 5(c^2-1)y = 0[(2+c)−3]x+5(c2−1)y=0 (c−1)x+5(c−1)(c+1)y=0. (c-1)x + 5(c-1)(c+1)y = 0.(c−1)x+5(c−1)(c+1)y=0.

    For c≠1c\neq 1c=1, divide by (c−1)(c-1)(c−1): x+5(c+1)y=0x + 5(c+1)y = 0x+5(c+1)y=0 x=−5(c+1)y.x = -5(c+1)y.x=−5(c+1)y.

  2. Substitute into 3x+5y=13x+5y=13x+5y=1: 3[−5(c+1)y]+5y=13[-5(c+1)y] + 5y = 13[−5(c+1)y]+5y=1 −15(c+1)y+5y=1-15(c+1)y + 5y = 1−15(c+1)y+5y=1 5[−3(c+1)+1]y=15[-3(c+1)+1]y = 15[−3(c+1)+1]y=1 5(−3c−2)y=15(-3c-2)y = 15(−3c−2)y=1 yc=−15(3c+2).y_c = -\frac{1}{5(3c+2)}.yc​=−5(3c+2)1​.

    Then xc=−5(c+1)yc=−5(c+1)(−15(3c+2))=c+13c+2.x_c = -5(c+1)y_c = -5(c+1)\left(-\frac{1}{5(3c+2)}\right)=\frac{c+1}{3c+2}.xc​=−5(c+1)yc​=−5(c+1)(−5(3c+2)1​)=3c+2c+1​.

  3. Take limits as c→1c\to 1c→1: h=lim⁡c→1xc=1+13+2=25,h = \lim_{c\to 1} x_c = \frac{1+1}{3+2} = \frac{2}{5},h=limc→1​xc​=3+21+1​=52​, k=lim⁡c→1yc=−15(5)=−125.k = \lim_{c\to 1} y_c = -\frac{1}{5(5)} = -\frac{1}{25}.k=limc→1​yc​=−5(5)1​=−251​.

    So the centre of the circle is (25,−125).\left(\frac{2}{5}, -\frac{1}{25}\right).(52​,−251​).

  4. Use that the circle passes through (2,0)(2,0)(2,0).

    Radius squared: r2=(2−25)2+(0+125)2r^2 = \left(2-\frac{2}{5}\right)^2 + \left(0+\frac{1}{25}\right)^2r2=(2−52​)2+(0+251​)2 =(85)2+(125)2= \left(\frac{8}{5}\right)^2 + \left(\frac{1}{25}\right)^2=(58​)2+(251​)2 =6425+1625=1600+1625=1601625.= \frac{64}{25} + \frac{1}{625} = \frac{1600+1}{625} = \frac{1601}{625}.=2564​+6251​=6251600+1​=6251601​.

    Hence the circle is (x−25)2+(y+125)2=1601625.\left(x-\frac{2}{5}\right)^2 + \left(y+\frac{1}{25}\right)^2 = \frac{1601}{625}.(x−52​)2+(y+251​)2=6251601​.

  5. Expand: x2−45x+425+y2+225y+1625=1601625.x^2 - \frac{4}{5}x + \frac{4}{25} + y^2 + \frac{2}{25}y + \frac{1}{625} = \frac{1601}{625}.x2−54​x+254​+y2+252​y+6251​=6251601​.

    Bring all terms to one side: x2+y2−45x+225y+(100625+1625−1601625)=0x^2+y^2-\frac{4}{5}x+\frac{2}{25}y+\left(\frac{100}{625}+\frac{1}{625}-\frac{1601}{625}\right)=0x2+y2−54​x+252​y+(625100​+6251​−6251601​)=0 x2+y2−45x+225y−1500625=0x^2+y^2-\frac{4}{5}x+\frac{2}{25}y-\frac{1500}{625}=0x2+y2−54​x+252​y−6251500​=0 x2+y2−45x+225y−125=0.x^2+y^2-\frac{4}{5}x+\frac{2}{25}y-\frac{12}{5}=0.x2+y2−54​x+252​y−512​=0.

    Multiply by 252525: 25x2+25y2−20x+2y−60=0.25x^2+25y^2-20x+2y-60=0.25x2+25y2−20x+2y−60=0.

  6. Match with options: This is Option B.

Therefore, the equation of the circle is 25x2+25y2−20x+2y−60=0.\boxed{25x^2+25y^2-20x+2y-60=0}. 25x2+25y2−20x+2y−60=0​.

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