JEE Mathematics Circle MCQs Set 04

Answer: (a) 3

Explanation:
1. A chord subtending \( \phi \) is at distance \( 2\cos\frac\phi2 \) from the centre. Same side is impossible (\( \cos t-\cos2t\le\frac98<\frac{\sqrt3+1}2 \)), so the chords are on opposite sides: \( 2\cos\frac\pi{2k}+2\cos\frac\pi k=\sqrt3+1 \).
2. This holds for \( k=3 \) (\( 2\cos30^\circ+2\cos60^\circ=\sqrt3+1 \)) and the left side increases with \( k \), so \( k=3 \) and \( [k]=3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( x^2+y^2=x+y \)

Explanation:
1. Putting \( y=x \): \( 2x^2-2x=0 \), so \( A(0,0) \) and \( B(1,1) \).
2. Circle on AB as diameter: \( x(x-1)+y(y-1)=0 \), i.e. \( x^2+y^2=x+y \).

Teacher's Note:
1. Find the end points first.
2. Use the diameter form (x - x1)(x - x2) + (y - y1)(y - y2) = 0.

Answer: (a) 2

Explanation:
1. Let \( CD=a,\ AB=2a,\ AD=2r \). Area \( \frac12(3a)(2r)=3ar=18 \), so \( ar=6 \).
2. Tangential quadrilateral: \( AB+CD=AD+BC \) with \( BC=\sqrt{a^2+4r^2} \), giving \( 3a-2r=\sqrt{a^2+4r^2} \) and \( 8a^2=12ar \), so \( a=\frac{3r}2 \).
3. Then \( \frac{3r^2}{2}=6 \) and \( r=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 5

Explanation:
1. On the y-axis the line gives \( -2y+6=0 \), so \( Q=(0,3) \).
2. The power of \( Q \) w.r.t. the circle is \( 0+9+0+18-2=25 \), so the tangent length is \( PQ=\sqrt{25}=5 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \left(\frac{1+\sqrt{31}}{4},\ \frac{7+\sqrt7}{4}\right] \)

Explanation:
1. P in B: \( (a-3)^2+(a-\tfrac12)^2\le 4 \) gives \( 2a^2-7a+\tfrac{25}{4}\le 4 \), i.e. \( 8a^2-28a+9\le 0 \), so \( \frac{7-\sqrt7}{4}\le a\le\frac{7+\sqrt7}{4} \).
2. P not in A: \( a^2+(a-\tfrac12)^2>4 \) gives \( 8a^2-4a-15>0 \), so \( a>\frac{1+\sqrt{31}}{4} \) or \( a<\frac{1-\sqrt{31}}{4} \).
3. Since \( \frac{1+\sqrt{31}}{4}\approx1.64 \) lies between \( 1.09 \) and \( 2.41 \), the common values are \( \frac{1+\sqrt{31}}{4}<a\le\frac{7+\sqrt7}{4} \).

Teacher's Note:
1. B - A means inside B but not inside A, so one condition is non-strict and the other strict.
2. Solve both quadratic inequalities and intersect the intervals.

Answer: (a) \( \left(\frac{\sqrt3}2,\frac32\right)\text{ and }(\sqrt3,0) \)

Explanation:
1. The centre is \( (\sqrt3,1) \) and \( \vec{CD}=\left(\frac{\sqrt3}2,\frac12\right) \) (direction \( 30^\circ \)).
2. The other two contact points are obtained by rotating \( \vec{CD} \) through \( \pm120^\circ \): \( \left(-\frac{\sqrt3}2,\frac12\right) \) and \( (0,-1) \).
3. So the points are \( \left(\frac{\sqrt3}{2},\frac32\right) \) and \( (\sqrt3,0) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \sqrt{PQ\cdot RS} \)

Explanation:
1. Take \( P(-r,0),R(r,0),Q(-r,q),S(r,s) \). The lines \( PS \) and \( RQ \) meet at \( X\left(\frac{r(q-s)}{q+s},\frac{qs}{q+s}\right) \).
2. On the circle: \( r^2[(q+s)^2-(q-s)^2]=q^2s^2 \), so \( 4r^2=qs \) and \( 2r=\sqrt{PQ\cdot RS} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( x^2+y^2-4x+2y=0 \)

Explanation:
1. The greatest distance from a diameter is the radius, so \( r=\sqrt5 \).
2. Centre \( (h,1-h) \) with \( h^2+(1-h)^2=5 \) (circle passes through the origin): \( h=2 \) or \( -1 \).
3. Centre \( (2,-1) \) gives \( x^2+y^2-4x+2y=0 \); the other centre \( (-1,2) \) gives \( x^2+y^2+2x-4y=0 \), which is not listed.

Teacher's Note:
1. Distance from the centre to any point of the circle is the radius.
2. The centre lies on the diameter.

Answer: (d) \( \pi\,2^{n-1}a^2 \)

Explanation:
1. The director circle has radius \( \sqrt2 \) times the original, so \( r_n=a\,2^{(n-1)/2} \).
2. Area between \( C_n \) and \( C_{n+1} \) = \( \pi(r_{n+1}^2-r_n^2)=\pi r_n^2=\pi\,2^{n-1}a^2 \).

Teacher's Note:
1. Radius multiplies by root 2 each time, so the radius squared doubles.
2. The annulus area equals the area of the inner circle.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. \( \vec{OQ}\cdot\vec{OR}=-12+12=0 \), so \( \angle QOR=\frac\pi2 \).
2. The angle at the circumference (on the major arc) is half the central angle: \( \angle QPR=\frac\pi4 \).

Teacher's Note:
1. Inscribed angle theorem: the angle at the circle is half the angle at the centre.
2. Use the dot product to find the central angle.

Answer: (a) 8

Explanation:
1. The centre of \( C \) lies on the perpendicular bisector at height \( h \): \( (1+r)^2=9+h^2 \).
2. The tangent length from \( P \) to \( C_1 \) is \( \sqrt{9-1} \) and to \( C \) is \( \sqrt{h^2-r^2} \); equal lengths give \( h^2-r^2=8 \).
3. Subtracting: \( 1+2r=17 \), so \( r=8 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( x^2+y^2-4x-10y+19=0 \)

Explanation:
1. The centre of the given circle is \( C(3,2) \). Since \( \angle PAC=\angle PBC=90^\circ \), the circumcircle of \( PAB \) has diameter \( PC \).
2. Equation: \( (x-1)(x-3)+(y-8)(y-2)=0 \), i.e. \( x^2+y^2-4x-10y+19=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( 1:4 \)

Explanation:
1. Solving \( x^2+8x=9 \): \( x=1 \), so \( P(1,2\sqrt2),\ Q(1,-2\sqrt2) \) and \( PQ=4\sqrt2 \).
2. Tangent to the circle at \( P \): \( x+2\sqrt2y=9 \), so \( R=(9,0) \). Tangent to the parabola: \( 2\sqrt2y=4(x+1) \), so \( S=(-1,0) \).
3. Heights from \( S \) and \( R \) to \( PQ \): 2 and 8, so the areas are in the ratio \( 2:8=1:4 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( y=\sqrt3x \) and \( y=0 \)

Explanation:
1. The contact points are \( \left(\frac{\sqrt3}2,\frac32\right) \) and \( (\sqrt3,0) \), and the tangent at each is perpendicular to the radius from \( (\sqrt3,1) \).
2. At \( (\sqrt3,0) \) the radius is vertical, so the tangent is \( y=0 \). At \( \left(\frac{\sqrt3}2,\frac32\right) \) the radius direction is \( \left(-\frac{\sqrt3}2,\frac12\right) \), giving \( -\sqrt3x+y=0 \), i.e. \( y=\sqrt3x \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( (-1,1) \)

Explanation:
1. The region is the upper half of the disc \( x^2+y^2<5 \) with \( y>0 \).
2. Inside the disc: \( a^2+(a+1)^2<5 \) gives \( a^2+a-2<0 \), so \( -2<a<1 \).
3. Above the axis: \( a+1>0 \), so \( a>-1 \). Together \( -1<a<1 \).

Teacher's Note:
1. Both the circle condition and the y > 0 condition must hold.
2. Intersect the intervals.

Answer: (d) \( 2ax+2by-(a^2+b^2+4)=0 \)

Explanation:
1. Let the circle be \( x^2+y^2+2gx+2fy+c=0 \). Orthogonality with \( x^2+y^2-4=0 \) gives \( c=4 \).
2. Passing through \( (a,b) \): \( a^2+b^2+2ga+2fb+4=0 \).
3. The centre \( (x,y)=(-g,-f) \) satisfies \( 2ax+2by-(a^2+b^2+4)=0 \).

Teacher's Note:
1. Orthogonal circles satisfy 2 g1 g2 + 2 f1 f2 = c1 + c2.
2. Replace g and f by minus the centre coordinates.

Answer: (b) Statement 1 is true and Statement 2 is false

Explanation:
1. The circle has centre \( (-3,5) \) and radius 2. \( L_1 \) passes through the centre when \( p=-6 \); then \( L_2 \) is \( 2x+3y-3=0 \) at distance \( \frac{6}{\sqrt{13}}<2 \) from the centre, so it is a chord: Statement 2 is false.
2. For \( p=0 \): \( L_1 \) is at distance \( \frac{6}{\sqrt{13}} \) (a chord) but \( L_2 \) is \( 2x+3y+3=0 \) which misses the centre: Statement 1 is true.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) (0, 3/2) or (5/4, 7/8)

Explanation:
1. The centre lies on the normal at \( (1,1) \): \( (1+2t,\ 1-t) \), with radius \( |t|\sqrt5 \).
2. Distance to \( 2x+y-4=0 \): \( \frac{|3t-1|}{\sqrt5} \) must equal \( |t|\sqrt5 \), so \( |3t-1|=5|t| \).
3. Then \( t=-\frac12 \) or \( t=\frac18 \), giving centres \( \left(0,\frac32\right) \) and \( \left(\frac54,\frac78\right) \).

Teacher's Note:
1. The radius to the point of contact is perpendicular to the tangent.
2. Equate distances from the centre to both tangents.

Answer: (d) Both are true and Statement 2 is a correct explanation of Statement 1

Explanation:
1. The director circle of \( x^2+y^2=169 \) is \( x^2+y^2=2\cdot169=338 \) (Statement 2 true).
2. For \( (17,7) \): \( 289+49=338 \), so the point is on the director circle and the tangents are perpendicular, which Statement 2 explains.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 5

Explanation:
1. Here \( OP=10 \). Tangent length \( L=\sqrt{100-r^2} \); half chord \( \frac{rL}{10} \); distance of chord from \( P \) is \( 10-\frac{r^2}{10} \).
2. Area \( =\frac{r(100-r^2)^{3/2}}{100} \). Setting the log-derivative to zero: \( \frac1r=\frac{3r}{100-r^2} \), so \( r^2=25 \), \( r=5 \).

Teacher's Note:
1. Express the triangle's area in terms of r and maximise.
2. Maximum at r = 5, where the circle radius is half the distance OP.

Answer: (d) \( (x-\sqrt3)^2+(y-1)^2=1 \)

Explanation:
1. The centre lies at distance 1 from \( PQ \) along the normal \( \left(\frac{\sqrt3}2,\frac12\right) \) towards the origin: \( \left(\frac{3\sqrt3}{2},\frac32\right)-\left(\frac{\sqrt3}{2},\frac12\right)=(\sqrt3,1) \).
2. So the circle is \( (x-\sqrt3)^2+(y-1)^2=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) a circle

Explanation:
1. For fixed \( A,B \), the centroid is \( G=\frac{A+B+P}{3} \).
2. As \( P \) traces a circle of radius \( r \), \( G \) traces a circle of radius \( \frac r3 \) centred at \( \frac{A+B}{3} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( 2\ \text{or}\ -\frac32 \)

Explanation:
1. Orthogonality: \( 2g_1g_2+2f_1f_2=c_1+c_2 \) with \( g_1=1,f_1=k,c_1=6 \) and \( g_2=0,f_2=k,c_2=k \).
2. So \( 2k^2=6+k \), i.e. \( 2k^2-k-6=0 \), giving \( k=2 \) or \( -\frac32 \).

Teacher's Note:
1. Use the orthogonality condition for general circles.
2. Solve the quadratic in k.

Answer: (d) 2

Explanation:
1. \( x^2y^2+1=x^2+y^2 \) gives \( (x^2-1)(y^2-1)=0 \), i.e. the lines \( x=\pm1,\ y=\pm1 \), a square.
2. The circle has centre \( (1,0) \) and radius 1.
3. Distance from centre to \( x=1 \) is 0 and to \( x=-1 \) is 2: not tangent.
4. Distance to \( y=1 \) and \( y=-1 \) is 1: both tangent (at \( (1,1) \) and \( (1,-1) \), which lie on the sides).
5. So 2 sides are touched.

Teacher's Note:
1. Factorise the joint equation first to get the four lines.
2. A line touches a circle when its distance from the centre equals the radius.

Answer: (d) 2

Explanation:
1. Sides: \( PQ=4\sqrt2 \), \( PR=QR=\sqrt{64+8}=6\sqrt2 \). Semi-perimeter \( s=8\sqrt2 \); area \( \frac12\cdot4\sqrt2\cdot8=16\sqrt2 \).
2. Inradius \( r=\frac{\Delta}{s}=\frac{16\sqrt2}{8\sqrt2}=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac{2b}{\sqrt{a^2-4b^2}} \)

Explanation:
1. The line already touches the first circle. Its distance from \( (a,0) \) must also equal \( b \): \( \frac{|ma-b\sqrt{1+m^2}|}{\sqrt{1+m^2}}=b \).
2. The non-trivial case is \( ma=2b\sqrt{1+m^2} \), so \( m^2a^2=4b^2(1+m^2) \) and \( m^2(a^2-4b^2)=4b^2 \).
3. For \( m>0 \): \( m=\frac{2b}{\sqrt{a^2-4b^2}} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) (ii) and (iii) only

Explanation:
1. A tangent to \( y^2=8ax \) is \( y=mx+\frac{2a}{m} \).
2. Distance from the origin must equal \( \sqrt2a \): \( \frac{4a^2}{m^2}=2a^2(1+m^2) \) gives \( m^4+m^2-2=0 \), so \( m=\pm1 \).
3. Tangents: \( y=x+2a \) and \( y=-x-2a \), i.e. (iii) and (ii).

Teacher's Note:
1. Use the slope form of a tangent to the parabola.
2. Then apply the condition for tangency to the circle.

Answer: (d) \( \frac34 \)

Explanation:
1. Take the centre as origin; vertices are at distance \( \sqrt2 \) from it. For a point at distance \( \rho \) from the centre, \( \sum PV^2=4\rho^2+4\cdot2 \).
2. For \( C_1 \) (\( \rho=1 \)): 12. For \( C_2 \) (\( \rho=\sqrt2 \)): 16.
3. Ratio \( \frac{12}{16}=\frac34 \).

Teacher's Note:
1. Sum of squared distances from a point to the vertices of a square depends only on its distance from the centre.
2. Radius of C1 is 1; radius of C2 is root 2.

Answer: (b) \( 3\sqrt3 \)

Explanation:
1. Points \( P(1,2\sqrt2),R(9,0),S(-1,0) \): \( RS=10 \), \( PS=\sqrt{12} \), \( PR=\sqrt{72} \); area \( \frac12\cdot10\cdot2\sqrt2=10\sqrt2 \).
2. Circumradius \( \frac{abc}{4\Delta}=\frac{10\sqrt{12}\sqrt{72}}{40\sqrt2}=3\sqrt3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.