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Circle question

2012 · Shift 1 · Q26
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  5. /2012 · Shift 1 · Q26

Circle question

2012 · Shift 1 · Q26

JEE AdvancedMathematicsCircleMCQ+4 / −1
The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4x - 5y = 20 to the circle x2 + y2=9{x^2}\, + \,{y^2} = 9x2+y2=9 is
  1. A
    20 (x2 + y2)−  36x  +  45y=020\,({x^2}\, + \,{y^2}) - \,\,36x\,\, + \,\,45y = 020(x2+y2)−36x+45y=0
  2. B
    20 (x2 + y2)+  36x  −  45y=020\,({x^2}\, + \,{y^2}) + \,\,36x\,\, - \,\,45y = 020(x2+y2)+36x−45y=0
  3. C
    36 (x2 + y2)−  20x  +  45y=036\,({x^2}\, + \,{y^2}) - \,\,20x\,\, + \,\,45y = 036(x2+y2)−20x+45y=0
  4. D
    36 (x2 + y2)+  20x  −  45y=036\,({x^2}\, + \,{y^2}) + \,\,20x\,\, - \,\,45y = 036(x2+y2)+20x−45y=0
View written solutionFree

Correct answer: A

Step-by-step Solution:

  1. Identify the given equations. The equation of the circle is S≡x2+y2−9=0S \equiv {x^2} + {y^2} - 9 = 0S≡x2+y2−9=0. Its center is at the origin (0,0)(0, 0)(0,0) and its radius is r=3r = 3r=3. The equation of the straight line is 4x−5y=204x - 5y = 204x−5y=20.

  2. Set up the coordinates. Let P(x1,y1)P(x_1, y_1)P(x1​,y1​) be any point on the given straight line from which tangents are drawn to the circle. Since P(x1,y1)P(x_1, y_1)P(x1​,y1​) lies on the line, it must satisfy the line's equation: 4x1−5y1=20...(1)4x_1 - 5y_1 = 20 \quad ... (1)4x1​−5y1​=20...(1) Let M(h,k)M(h, k)M(h,k) be the midpoint of the chord of contact of the tangents drawn from P(x1,y1)P(x_1, y_1)P(x1​,y1​). We need to find the locus of M(h,k)M(h, k)M(h,k).

  3. Find the equation of the chord of contact. The equation of the chord of contact of tangents drawn from an external point P(x1,y1)P(x_1, y_1)P(x1​,y1​) to the circle x2+y2=9x^2 + y^2 = 9x2+y2=9 is given by the formula T=0T = 0T=0. xx1+yy1=9...(2)xx_1 + yy_1 = 9 \quad ... (2)xx1​+yy1​=9...(2)

  4. Find the equation of the chord with a given midpoint. The equation of a chord of the circle x2+y2=9x^2 + y^2 = 9x2+y2=9 whose midpoint is M(h,k)M(h, k)M(h,k) is given by the formula T=S1T = S_1T=S1​. Here, T=hx+ky−9T = hx + ky - 9T=hx+ky−9 and S1=h2+k2−9S_1 = h^2 + k^2 - 9S1​=h2+k2−9. So, the equation is: hx+ky−9=h2+k2−9hx + ky - 9 = h^2 + k^2 - 9hx+ky−9=h2+k2−9 hx+ky=h2+k2...(3)hx + ky = h^2 + k^2 \quad ... (3)hx+ky=h2+k2...(3)

  5. Equate the two representations of the chord. Equations (2) and (3) both represent the same line (the chord of contact). Therefore, the coefficients of xxx, yyy, and the constant terms must be proportional. Comparing equation (2), xx1+yy1=9x x_1 + y y_1 = 9xx1​+yy1​=9, and equation (3), xh+yk=h2+k2x h + y k = h^2 + k^2xh+yk=h2+k2, we get: x1h=y1k=9h2+k2\frac{x_1}{h} = \frac{y_1}{k} = \frac{9}{h^2 + k^2}hx1​​=ky1​​=h2+k29​

  6. Express x1x_1x1​ and y1y_1y1​ in terms of hhh and kkk. From the proportionality, we can express x1x_1x1​ and y1y_1y1​ as: x1=9hh2+k2x_1 = \frac{9h}{h^2 + k^2}x1​=h2+k29h​ y1=9kh2+k2y_1 = \frac{9k}{h^2 + k^2}y1​=h2+k29k​

  7. Use the condition that P(x1,y1)P(x_1, y_1)P(x1​,y1​) lies on the line. Substitute the expressions for x1x_1x1​ and y1y_1y1​ into the equation of the line (1): 4(9hh2+k2)−5(9kh2+k2)=204\left(\frac{9h}{h^2 + k^2}\right) - 5\left(\frac{9k}{h^2 + k^2}\right) = 204(h2+k29h​)−5(h2+k29k​)=20

  8. Simplify to find the locus. 36hh2+k2−45kh2+k2=20\frac{36h}{h^2 + k^2} - \frac{45k}{h^2 + k^2} = 20h2+k236h​−h2+k245k​=20 36h−45kh2+k2=20\frac{36h - 45k}{h^2 + k^2} = 20h2+k236h−45k​=20 36h−45k=20(h2+k2)36h - 45k = 20(h^2 + k^2)36h−45k=20(h2+k2) Rearranging the terms, we get: 20(h2+k2)−36h+45k=020(h^2 + k^2) - 36h + 45k = 020(h2+k2)−36h+45k=0

  9. Generalize the locus. To find the locus of the midpoint M(h,k)M(h, k)M(h,k), we replace (h,k)(h, k)(h,k) with (x,y)(x, y)(x,y): 20(x2+y2)−36x+45y=020(x^2 + y^2) - 36x + 45y = 020(x2+y2)−36x+45y=0

  10. Compare with the given options. The derived locus is 20(x2+y2)−36x+45y=020(x^2 + y^2) - 36x + 45y = 020(x2+y2)−36x+45y=0, which matches option A.

Conclusion

The correct option is A.

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