Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Circle question

2021 · 22 Jul · Shift 2 · Q33
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Circle
  5. /2021 · 22 Jul · Shift 2 · Q33

Circle question

2021 · 22 Jul · Shift 2 · Q33

JEE MainMathematicsCircleMCQ+4 / −1
Let the circle S : 36x2 + 36y2 −-− 108x + 120y + C = 0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x −-− 2y = 4 and 2x −-− y = 5 lies inside the circle S, then :
  1. A
    259<C<133{{25} \over 9} \lt C \lt {{13} \over 3}925​<C<313​
  2. B
    100 < C < 165
  3. C
    81 < C < 156
  4. D
    100 < C < 156
View written solutionFree

Correct answer: D

  1. Write the circle in standard form

Given 36x2+36y2−108x+120y+C=036x^2+36y^2-108x+120y+C=036x2+36y2−108x+120y+C=0 Divide by 363636: x2+y2−3x+103y+C36=0x^2+y^2-3x+\frac{10}{3}y+\frac{C}{36}=0x2+y2−3x+310​y+36C​=0

Compare with x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 So, 2g=−3⇒g=−32,2f=103⇒f=532g=-3\Rightarrow g=-\frac{3}{2}, \qquad 2f=\frac{10}{3}\Rightarrow f=\frac{5}{3}2g=−3⇒g=−23​,2f=310​⇒f=35​

Hence center is (−g,−f)=(32,−53)(-g,-f)=\left(\frac{3}{2},-\frac{5}{3}\right)(−g,−f)=(23​,−35​)

Radius satisfies r2=g2+f2−c=94+259−C36r^2=g^2+f^2-c=\frac{9}{4}+\frac{25}{9}-\frac{C}{36}r2=g2+f2−c=49​+925​−36C​ Taking LCM 363636, r2=81+100−C36=181−C36r^2=\frac{81+100-C}{36}=\frac{181-C}{36}r2=3681+100−C​=36181−C​

So, r=181−C6r=\frac{\sqrt{181-C}}{6}r=6181−C​​

  1. Condition: circle neither intersects nor touches the coordinate axes

For a circle to neither touch nor intersect an axis, its radius must be less than the perpendicular distance of the center from that axis.

  • Distance from center to xxx-axis: ∣−53∣=53\left| -\frac{5}{3} \right|=\frac{5}{3}​−35​​=35​
  • Distance from center to yyy-axis: ∣32∣=32\left| \frac{3}{2} \right|=\frac{3}{2}​23​​=23​

Therefore, r<53andr<32r<\frac{5}{3} \quad \text{and} \quad r<\frac{3}{2}r<35​andr<23​ The stricter condition is r<32r<\frac{3}{2}r<23​

So, 181−C6<32\frac{\sqrt{181-C}}{6}<\frac{3}{2}6181−C​​<23​ 181−C<9\sqrt{181-C}<9181−C​<9 181−C<81181-C<81181−C<81 C>100C>100C>100

Also, for a real circle, r2>0⇒181−C>0⇒C<181r^2>0 \Rightarrow 181-C>0 \Rightarrow C<181r2>0⇒181−C>0⇒C<181

From this condition alone: 100<C<181100<C<181100<C<181

  1. Find the intersection point of the lines

Given lines: x−2y=4x-2y=4x−2y=4 2x−y=52x-y=52x−y=5

Solve them.

From first equation, x=4+2yx=4+2yx=4+2y Substitute into second: 2(4+2y)−y=52(4+2y)-y=52(4+2y)−y=5 8+4y−y=58+4y-y=58+4y−y=5 8+3y=58+3y=58+3y=5 3y=−3⇒y=−13y=-3\Rightarrow y=-13y=−3⇒y=−1 Then x=4+2(−1)=2x=4+2(-1)=2x=4+2(−1)=2

So the point is (2,−1)(2,-1)(2,−1)

  1. Condition: this point lies inside the circle

Distance from center (32,−53)\left(\frac{3}{2},-\frac{5}{3}\right)(23​,−35​) to point (2,−1)(2,-1)(2,−1):

d2=(2−32)2+(−1+53)2d^2=\left(2-\frac{3}{2}\right)^2+\left(-1+\frac{5}{3}\right)^2d2=(2−23​)2+(−1+35​)2 =(12)2+(23)2=\left(\frac{1}{2}\right)^2+\left(\frac{2}{3}\right)^2=(21​)2+(32​)2 =14+49=\frac{1}{4}+\frac{4}{9}=41​+94​ =9+1636=2536=\frac{9+16}{36}=\frac{25}{36}=369+16​=3625​

Thus, d=56d=\frac{5}{6}d=65​

For the point to lie inside the circle, r>dr>dr>d 181−C6>56\frac{\sqrt{181-C}}{6}>\frac{5}{6}6181−C​​>65​ 181−C>5\sqrt{181-C}>5181−C​>5 181−C>25181-C>25181−C>25 C<156C<156C<156

  1. Combine both conditions

From axis condition: C>100C>100C>100 From inside-point condition: C<156C<156C<156

Therefore, 100<C<156100<C<156100<C<156

  1. Check options
  • A: 259<C<133\frac{25}{9}<C<\frac{13}{3}925​<C<313​, false
  • B: 100<C<165100<C<165100<C<165, too wide
  • C: 81<C<15681<C<15681<C<156, too wide
  • D: 100<C<156100<C<156100<C<156, correct

Hence the correct option is: D\boxed{D}D​

PreviousNext

More from Circle

  • If one of the diameters of the circle x2 + y2 - 2x - 6y + 6 = 0 is a chord of another circle 'C', whose center is at (2, 1), then its radius is ​.2021 · Numerical
  • Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point (− 5, 0). If the locus of the point P is a circle of radius r, then 4r2 is equal to ​2021 · Numerical
  • If a line along a chord of the circle 4x2 + 4y2 + 120x + 675 = 0, passes through the point (− 30, 0) and is tangent to the parabola y2 = 30x, then the length of this chord is :2021 · MCQ
  • The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d2 is equal to ​.2021 · Numerical
  • A circle C touches the line x = 2y at the point (2, 1) and intersects the circle C1 : x2 + y2 + 2y − 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :2021 · MCQ
  • In the circle given below, let OA = 1 unit, OB = 13 unit and PQ ⊥ OB. Then, the area of the triangle PQB (in square units) is : Includes diagram2021 · MCQ
  • Let A(1, 4) and B(1, − 5) be two points. Let P be a point on the circle (x − 1)2 + (y − 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :2021 · MCQ
  • If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2 + y2 = 1 is a circle of radius r, then r is equal to :2021 · MCQ