JEE Mathematics Circles MCQs Set 12

Answer: (c) \( xy=0 \)

Explanation:
1. Tangent \( y=mx+a\sqrt{1+m^2} \) through \( P(h,k) \) gives \( (h^2-a^2)m^2-2hkm+(k^2-a^2)=0 \).
2. \( \cot\alpha+\cot\beta=\frac{m_1+m_2}{m_1m_2}=\frac{2hk}{k^2-a^2}=0 \), so \( hk=0 \).
3. The locus is \( xy=0 \).

Teacher's Note:
1. Form the quadratic in the slope m using the tangent condition.
2. cot α + cot β = (m₁ + m₂)/(m₁m₂).

Answer: (b) (1, 2)

Explanation:
1. The centre is \( (2,-1) \) and \( r=\sqrt{10} \). A line \( x-3y+c=0 \) is tangent when \( \frac{|5+c|}{\sqrt{10}}=\sqrt{10} \), so \( c=5 \) or \( c=-15 \).
2. The points of contact are the feet of the perpendiculars from the centre: \( (1,2) \) for \( c=5 \) and \( (3,-4) \) for \( c=-15 \).
3. Only \( (1,2) \) is in the first quadrant.

Teacher's Note:
1. Two tangents are parallel to the given line, one on each side of the centre.
2. Pick the contact point lying in the first quadrant.

Answer: (d) \( \left(\frac12,\pm\sqrt2\right) \)

Explanation:
1. The centre lies on the perpendicular bisector of the chord: \( \left(\frac12,k\right) \), and the radius is its distance from the origin, \( r=\sqrt{\frac14+k^2} \).
2. The circle passes through the origin, which is inside \( x^2+y^2=9 \), so it touches internally: distance of centres \( =3-r \), i.e. \( r=3-r \).
3. \( r=\frac32 \), so \( k^2=\frac94-\frac14=2 \) and the centre is \( \left(\frac12,\pm\sqrt2\right) \).

Teacher's Note:
1. The centre lies on the perpendicular bisector of the common chord.
2. For internal contact, distance between centres = difference of radii.

Answer: (a) \( (x_1^2+y_1^2+2gx_1+2fy_1+c)^{1/2} \)

Explanation:
1. The length of the tangent is \( \sqrt{S_1} \) where \( S_1 \) is the circle's expression at \( (x_1,y_1) \).
2. This is option (a).

Teacher's Note:
1. It is the square root of the power of the point.
2. Option (c) is the distance to the centre, not the tangent length.

Answer: (c) 40

Explanation:
1. \( S_1=9+16-12+24+3=40 \).
2. The square of the tangent length is \( S_1=40 \).

Teacher's Note:
1. The square of the tangent length is S₁ itself.
2. Be careful with the sign of −6y when y = −4.

Answer: (d) \( \sqrt{\frac52} \)

Explanation:
1. Divide by 2: \( x^2+y^2+\frac x2-\frac y2+\frac52=0 \).
2. \( S_1 \) at the origin is the constant term \( \frac52 \), so the length is \( \sqrt{\frac52} \).

Teacher's Note:
1. Make the coefficients of x² and y² equal to 1 first.
2. At the origin S₁ equals the constant term.

Answer: (b) 2

Explanation:
1. The circle is \( x^2+y^2-\frac32x+2y=0 \).
2. \( S_1=4+1-3+2=4 \), so \( PA=\sqrt4=2 \).

Teacher's Note:
1. PA is the length of the tangent from P.
2. Divide by 2 before substituting.

Answer: (b) \( \left(-2,-\frac52\right) \)

Explanation:
1. Equal tangent lengths mean the point is on the radical axes.
2. First two circles: \( -1=8x+15 \) gives \( x=-2 \). First and third: \( -1=10y+24 \) gives \( y=-\frac52 \).
3. The point is \( \left(-2,-\frac52\right) \).

Teacher's Note:
1. The point is the radical centre.
2. Equate the S₁ values pairwise.

Answer: (b) (-7, 8)

Explanation:
1. Let \( S_1,S_2 \) be the two expressions. Then \( 3S_1=2S_2 \).
2. \( x^2+y^2+14x-16y+28=0 \).
3. The centre is \( (-7,8) \).

Teacher's Note:
1. Square of tangent length = S at the point.
2. Collect terms to read off the centre (−g, −f).

Answer: (c) \( a^2,b^2,c^2 \) are in A.P.

Explanation:
1. The squares are \( r^2-a^2,\ r^2-b^2,\ r^2-c^2 \) with \( r^2=x^2+y^2 \).
2. These are in A.P. exactly when \( -a^2,-b^2,-c^2 \) are in A.P., i.e. \( a^2,b^2,c^2 \) are in A.P.

Teacher's Note:
1. Subtracting a constant keeps an A.P. an A.P.
2. Use 2·middle = sum of the extremes.

Answer: (b) \( (h^2-r^2)x-2rhy=0,\ x=0 \)

Explanation:
1. The centre is \( (r,h) \) and \( r_0^2=r^2+h^2-h^2=r^2 \), so the radius is \( r \); the circle touches the \( y\text{-axis} \), so \( x=0 \) is one tangent.
2. For \( y=mx \): \( \frac{|mr-h|}{\sqrt{1+m^2}}=r \) gives \( -2mrh=r^2-h^2 \), so \( m=\frac{h^2-r^2}{2rh} \).
3. The other tangent is \( (h^2-r^2)x-2rhy=0 \).

Teacher's Note:
1. Check whether the circle touches an axis.
2. Use distance from centre = radius for the second line.

Answer: (a) \( 2x-y+1=0,\ x+2y-2=0 \)

Explanation:
1. The centre is \( (1,-2) \) and \( r=\sqrt5 \). Take \( y=mx+1 \).
2. \( \frac{|m+3|}{\sqrt{1+m^2}}=\sqrt5 \) gives \( 2m^2-3m-2=0 \), so \( m=2 \) or \( m=-\frac12 \).
3. The lines are \( 2x-y+1=0 \) and \( x+2y-2=0 \).

Teacher's Note:
1. Write the tangent through (0, 1) as y = mx + 1.
2. Distance of the centre from the line equals the radius.

Answer: (a) \( 7x^2+23y^2+30xy+66x+50y-73=0 \)

Explanation:
1. \( S_1=36+25-12-20+3=32 \) and \( T=6x-5y-(x+6)+2(y-5)+3=5x-3y-13 \).
2. \( SS_1=T^2 \): \( 32(x^2+y^2-2x+4y+3)=(5x-3y-13)^2 \).
3. Expanding gives \( 7x^2+23y^2+30xy+66x+50y-73=0 \).

Teacher's Note:
1. Use SS₁ = T².
2. Expand and collect like terms carefully.

Answer: (c) \( x+2y=\pm2\sqrt5 \)

Explanation:
1. Tangents \( x+2y=k \) need \( \frac{|k|}{\sqrt5}=2 \).
2. So \( k=\pm2\sqrt5 \).

Teacher's Note:
1. Parallel lines share the x and y coefficients.
2. Distance from the origin must equal 2.

Answer: (c) \( 24x-7y+125=0 \)

Explanation:
1. A line through \( (-2,11) \) is tangent when its distance from the origin is 5.
2. \( 24x-7y+125=0 \) passes through the point (\( -48-77+125=0 \)) and has distance \( \frac{125}{25}=5 \).
3. The other options either miss the point or are not at distance 5.

Teacher's Note:
1. Test each option for both conditions.
2. Tangent: distance from the origin = 5.

Answer: (b) \( \sin^{-1}\frac45 \)

Explanation:
1. The centre is \( (-2,-1) \), \( r=3 \), and \( d^2=9+\frac94=\frac{45}4 \).
2. \( \sin\frac\theta2=\frac3d=\frac2{\sqrt5} \), so \( \cos\theta=1-2\cdot\frac45=-\frac35 \) and \( \sin\theta=\frac45 \). The angle between the tangents at the point is obtuse (about \( 127^\circ \)).
3. The acute angle between the two lines is the supplement: \( \sin^{-1}\frac45 \) (about \( 53^\circ \)).

Teacher's Note:
1. Use sin(θ/2) = r/d and then find sin θ and cos θ.
2. The acute angle between two lines is 180° minus the obtuse angle.

Answer: (d) \( \frac\pi2 \)

Explanation:
1. The centre is \( (7,-1) \), \( r=5 \), and the distance from the origin is \( \sqrt{50} \).
2. \( \sin\frac\theta2=\frac5{\sqrt{50}}=\frac1{\sqrt2} \), so \( \theta=\frac\pi2 \).

Teacher's Note:
1. Half the angle between tangents has sine r/d.
2. Here d = 5√2.

Answer: (c) \( \frac{192}{25} \)

Explanation:
1. For a circle of radius \( a \) and external point at distance \( d \), with tangent length \( L \), the area is \( \frac{aL^3}{d^2} \).
2. Here \( a=3 \), \( d=5 \), \( L=\sqrt{25-9}=4 \).
3. Area \( =\frac{3\cdot64}{25}=\frac{192}{25} \).

Teacher's Note:
1. The chord of contact is at distance a²/d from the centre.
2. Area = (1/2) × chord × height.

Answer: (a) 75 sq. units

Explanation:
1. The centre is \( (1,2) \) and \( r=5 \). \( S_1=256+49-32-28-20=225 \), so the tangent length is 15.
2. The quadrilateral is two right triangles with legs 5 and 15: area \( =2\cdot\frac12\cdot5\cdot15=75 \).

Teacher's Note:
1. Split the quadrilateral into triangles PQC and PRC.
2. Each is right-angled at the point of contact.

Answer: (a) (10, 0)

Explanation:
1. Let \( P=(p,0) \). A tangent \( y=m(x-p) \) needs \( m^2=\frac{50}{p^2-50} \), so the \( y \)-intercepts are \( \pm p\sqrt{\frac{50}{p^2-50}} \).
2. Area \( =\frac12\cdot2p\sqrt{\frac{50}{p^2-50}}\cdot p=\frac{\sqrt{50}\,p^2}{\sqrt{p^2-50}} \).
3. With \( u=p^2 \), \( \frac{u}{\sqrt{u-50}} \) is minimum at \( u=100 \), so \( p=10 \).

Teacher's Note:
1. Find the y-intercepts of the two tangents in terms of p.
2. Minimise the area as a function of p².

Answer: (a) \( 2x+9y-30=0 \)

Explanation:
1. The centre is \( \left(\frac32,3\right) \). The normal passes through it and \( (-3,4) \).
2. Slope \( =\frac{3-4}{\frac32+3}=-\frac29 \), so \( y-4=-\frac29(x+3) \), i.e. \( 2x+9y-30=0 \).

Teacher's Note:
1. A normal to a circle passes through the centre.
2. Find the slope from the two points.

Answer: (b) \( x^2+y^2-2x-2y-11=0 \)

Explanation:
1. A normal passes through the centre, so the two points are ends of a diameter.
2. Centre \( =(1,1) \), \( r^2=(3-1)^2+(4-1)^2=13 \).
3. Equation: \( (x-1)^2+(y-1)^2=13 \), i.e. \( x^2+y^2-2x-2y-11=0 \).

Teacher's Note:
1. The normal meets the circle again at the other end of the diameter.
2. The centre is the midpoint of the two points.