JEE Mathematics Circles MCQs Set 20

Answer: (d) \( 2x+3y=0 \)

Explanation:
1. The circle through the three points is \( x^2+y^2-4x-6y=0 \), with centre \( (2,3) \); it passes through the origin.
2. So the tangent at the origin is perpendicular to the radius direction \( (2,3) \): \( 2x+3y=0 \).

Teacher's Note:
1. Find the circle through the three points.
2. Check whether the origin lies on it.

Answer: (a) \( \tfrac1{\sqrt3} \)

Explanation:
1. The inradius is \( R=\tfrac{6\sqrt3}6=\sqrt3 \).
2. The small circle near a vertex lies inside the angle of \( 60^\circ \): \( \dfrac{R-r}{R+r}=\sin30^\circ=\tfrac12 \), so \( r=\tfrac R3=\dfrac1{\sqrt3} \).

Teacher's Note:
1. The centres lie on the angle bisector at the vertex.
2. Use the sine of the half-angle for the ratio of radii.

Answer: (a) \( \tfrac75 \)

Explanation:
1. The circle has centre \( (3,-4) \) and radius 5.
2. Distance of the centre from the line: \( \dfrac{|9-16-25|}5=\tfrac{32}5 \). Shortest distance \( =\tfrac{32}5-5=\tfrac75 \).

Teacher's Note:
1. Find the centre and radius first.
2. Distance from line to circle = distance from centre minus radius.

Answer: (b) \( 6\sqrt{26} \)

Explanation:
1. Let \( r=AB \). Power of C: \( CD\cdot CB=CE\cdot CF \) with \( CE=AC-r \) and \( CF=AC+r \): \( 16\cdot36=AC^2-r^2 \).
2. Right angle at A: \( AC^2+r^2=36^2 \). Adding: \( 2AC^2=576+1296 \), so \( AC^2=936 \) and \( AC=6\sqrt{26} \).

Teacher's Note:
1. Use the power of the point C with respect to the circle.
2. Combine with Pythagoras in triangle ABC.

Answer: (b) \( x^2+8x+(y-3)^2=0 \)

Explanation:
1. B is the mid-point of AM. If \( M=(x,y) \) then \( B=\left(\tfrac x2,\tfrac{y+3}2\right) \) lies on the circle.
2. \( \tfrac{x^2}4+2x+\left(\tfrac{y-3}2\right)^2=0 \), i.e. \( x^2+8x+(y-3)^2=0 \).

Teacher's Note:
1. Write B in terms of M and A.
2. Substitute B in the circle equation.

Answer: (b) \( x^2-8y^2-54x+729=0 \)

Explanation:
1. The chord of contact from \( (h,k) \) is \( hx+ky=81 \); matching \( x=3 \) gives the point \( (27,0) \).
2. Pair of tangents: \( SS_1=T^2 \): \( (x^2+y^2-81)(729-81)=(27x-81)^2 \), i.e. \( 8(x^2+y^2-81)=9(x-3)^2 \).
3. Simplifying: \( x^2-8y^2-54x+729=0 \).

Teacher's Note:
1. Find the external point from the chord of contact.
2. Use SS1 = T^2.

Answer: (a) \( x^2+y^2-5x-4y+6=0 \)

Explanation:
1. The mid-point M satisfies \( CM\perp PM \) with \( C=(2,0) \) and \( P=(3,4) \).
2. So M lies on the circle with diameter CP: \( (x-3)(x-2)+(y-4)y=0 \), i.e. \( x^2+y^2-5x-4y+6=0 \).

Teacher's Note:
1. The line from the centre to a chord's mid-point is perpendicular to the chord.
2. The locus is a circle on the segment from the centre to the given point.

Answer: (c) \( 9x-10y+11=0 \)

Explanation:
1. A circle cutting two circles orthogonally has its centre on their radical axis.
2. Subtracting the equations: \( 9x-10y+11=0 \).

Teacher's Note:
1. The centre of an orthogonal circle has equal power with respect to both circles.
2. Compute S1 - S2.

Answer: (a) \( 2x-5y+11=0 \)

Explanation:
1. The centre lies on the perpendicular to the chord through its mid-point \( (2,3) \).
2. The chord has slope \( -\tfrac52 \), so the perpendicular has slope \( \tfrac25 \): \( y-3=\tfrac25(x-2) \), i.e. \( 2x-5y+11=0 \).

Teacher's Note:
1. The perpendicular from the centre bisects a chord.
2. Take the line through (2,3) perpendicular to the chord.

Answer: (b) \( (x+2)^2+(y-3)^2=6.25 \)

Explanation:
1. The circle has centre \( (-2,3) \) and radius 5. An inscribed angle of \( \tfrac\pi3 \) means a central angle of \( \tfrac{2\pi}3 \).
2. The mid-point of such a chord is at distance \( 5\cos\tfrac\pi3=2.5 \) from the centre: \( (x+2)^2+(y-3)^2=6.25 \).

Teacher's Note:
1. Double the inscribed angle to get the central angle.
2. Distance of a chord's mid-point from the centre is r cos(half the central angle).

Answer: (d) none of the points lies outside S

Explanation:
1. Evaluate S at each point: \( P \): \( 4+1-10+2-5=-8 \); \( Q \): \( -5 \); \( R \): \( 16+9-20-6-5=-6 \).
2. All values are negative, so all three points lie inside the circle.

Teacher's Note:
1. A point is inside the circle when S(x1, y1) is negative.
2. Substitute each point.

Answer: (b) \( \tfrac\pi4 \)

Explanation:
1. Centres \( (1,0) \) and \( (0,2) \), so \( d^2=5 \); radii squared 10 and 5.
2. \( \cos\theta=\dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}=\dfrac{10}{2\sqrt{50}}=\dfrac1{\sqrt2} \), so \( \theta=\tfrac\pi4 \).

Teacher's Note:
1. Use the cosine rule in the triangle formed by the two centres and a common point.
2. The angle between the circles is the angle between the radii.

Answer: (c) \( a^2-4ab+b^2=0 \)

Explanation:
1. A circle touching both axes has centre \( (\rho,\rho) \) and radius \( \rho \); through P: \( \rho^2-2(a+b)\rho+a^2+b^2=0 \) with roots \( \rho_1,\rho_2 \).
2. Orthogonality: \( 2(\rho_1-\rho_2)^2=\rho_1^2+\rho_2^2 \), i.e. \( (\rho_1+\rho_2)^2=6\rho_1\rho_2 \).
3. So \( 4(a+b)^2=6(a^2+b^2) \), giving \( a^2-4ab+b^2=0 \).

Teacher's Note:
1. The two radii are the roots of a quadratic.
2. Orthogonal circles satisfy d^2 = r1^2 + r2^2.

Answer: (b) 11

Explanation:
1. The nearest vertices are at the \( 120^\circ \) angles: \( r=1\cdot\sin60^\circ=\tfrac{\sqrt3}2 \); the other two vertices are at distance \( \tfrac{r}{\sin30^\circ}=\sqrt3 \).
2. Take the centre as the origin, vertices \( (\pm1,0) \), \( (0,\pm\sqrt3) \). Then the sum is \( 4|P|^2+\sum|V|^2=4\cdot\tfrac34+(1+1+3+3)=11 \).

Teacher's Note:
1. Place the centre at the origin with the diagonals along the axes.
2. Use the sum of squared distances formula.

Answer: (d) \( (-6,-7) \)

Explanation:
1. Centre \( (-8,-6) \). The line is \( 2x-y+5=0 \). The point of contact is the foot of the perpendicular from the centre.
2. \( 2(-8)+6+5=-5 \), so the foot is \( (-8,-6)-\tfrac{-5}{5}(2,-1)=(-6,-7) \).

Teacher's Note:
1. The point of contact is the foot of the perpendicular from the centre.
2. Use the foot formula.

Answer: (c) \( x^2+y^2-2x-1=0 \)

Explanation:
1. The normal line through \( (4,3) \) and \( (2,1) \) is \( y=x-1 \) and it passes through the centre.
2. Intersect with \( 2x-y-2=0 \): centre \( (1,0) \). Radius\( ^2 \) \( =(2-1)^2+1^2=2 \).
3. Circle: \( (x-1)^2+y^2=2 \), i.e. \( x^2+y^2-2x-1=0 \).

Teacher's Note:
1. A normal passes through the centre.
2. The centre is the common point of two lines through it.

Answer: (b) \( 3\sqrt2 \)

Explanation:
1. Centre \( (1,1) \), \( r=3 \). Distance from \( (4,4) \): \( d=3\sqrt2 \); tangent length \( L=\sqrt{18-9}=3 \).
2. Chord of contact \( =\dfrac{2Lr}{d}=\dfrac{18}{3\sqrt2}=3\sqrt2 \).

Teacher's Note:
1. Chord of contact length is 2Lr/d.
2. Find the tangent length from the power of the point.

Answer: (a) \( x^2+y^2-10x+4y+28=0 \)

Explanation:
1. The given circle has centre \( (1,2) \) and radius 1. Reflection in \( x-y=3 \) sends \( (x,y)\to(y+3,x-3) \), so the new centre is \( (5,-2) \).
2. Circle: \( (x-5)^2+(y+2)^2=1 \), i.e. \( x^2+y^2-10x+4y+28=0 \).

Teacher's Note:
1. Reflect the centre; the radius is unchanged.
2. Reflection in x - y = 3 swaps coordinates with a shift.

Answer: (d) \( y^2-10x-6y+14=0 \)

Explanation:
1. The given circle has centre \( (3,3) \), radius 2. A circle with centre \( (h,k) \) touching the y-axis has radius \( h \).
2. External contact: \( (h-3)^2+(k-3)^2=(h+2)^2 \), giving \( k^2-6k-10h+14=0 \).
3. Locus: \( y^2-10x-6y+14=0 \).

Teacher's Note:
1. The radius of a circle touching the y-axis equals the x-coordinate of its centre.
2. External tangency: distance between centres equals the sum of radii.

Answer: (a) \( 16(x^2+y^2)-48x+16y+31=0 \)

Explanation:
1. Divide by 4: centre \( \left(\tfrac32,-\tfrac12\right) \) and \( r^2=\tfrac94+\tfrac14-\tfrac14=\tfrac94 \), \( r=\tfrac32 \).
2. The mid-point is at distance \( r\cos\tfrac\pi3=\tfrac34 \) from the centre.
3. \( \left(x-\tfrac32\right)^2+\left(y+\tfrac12\right)^2=\tfrac9{16} \) gives \( 16(x^2+y^2)-48x+16y+31=0 \).

Teacher's Note:
1. Half of the central angle makes the triangle from the centre to the mid-point.
2. Distance from the centre is r cos(60 degrees).

Answer: (b) 2

Explanation:
1. The first equation is the pair of lines \( y=\pm x \) meeting at the origin; the second is a unit circle with centre \( (A,0) \).
2. Three common points occur when the circle passes through the origin: \( A=\pm1 \) (points \( (0,0) \) and \( (1,\pm1) \) for \( A=1 \)). If instead it is tangent to both lines (\( A=\pm\sqrt2 \)) there are only 2 points.
3. So 2 values of A.

Teacher's Note:
1. Interpret the system geometrically.
2. Check the case where the circle passes through the intersection of the lines.

Answer: (b) \( \left(-\tfrac1{10},\tfrac1{10}\right) \)

Explanation:
1. Power \( =a^2+b^2-2(t^2-3t+1)a-2(t^2+2t)b+t \). Coefficient of \( t^2 \): \( -2a-2b=0 \), so \( b=-a \).
2. Coefficient of \( t \): \( 6a-4b+1=0 \), so \( 10a+1=0 \), giving \( a=-\tfrac1{10} \), \( b=\tfrac1{10} \).

Teacher's Note:
1. Collect the power as a polynomial in t.
2. Set the coefficients of t^2 and t to zero.

Answer: (a) \( \tfrac75 \)

Explanation:
1. In the right triangle formed by O, A and the point of contact, \( \sin\varphi=\tfrac35 \) where \( \varphi \) is the angle between the tangent and AO; so \( \cos\varphi=\tfrac45 \).
2. The chord \( AD=2\cdot5\cos\varphi=8 \), so \( D=(0,5)+8\left(\tfrac35,-\tfrac45\right)=\left(\tfrac{24}5,-\tfrac75\right) \). By symmetry DE is the line \( y=-\tfrac75 \).
3. So \( OF=\tfrac75 \).

Teacher's Note:
1. Find the angle between each tangent and the line AO.
2. A chord from A makes length 2R cos(angle with the diameter).

Answer: (d) an arc of a circle with OP as diameter

Explanation:
1. The line from the centre to the mid-point of a chord is perpendicular to the chord, so \( \angle OSP=90^\circ \).
2. By Thales' theorem S lies on the circle with diameter OP.

Teacher's Note:
1. OS is perpendicular to QR.
2. A right angle subtended by OP means S lies on the circle on OP as diameter.

Answer: (d) \( PS^2 \)

Explanation:
1. By the power of a point, \( PQ\cdot PR=d^2-r^2=PT^2 \).
2. Also \( PQ\cdot PR=(PS-QS)(PS+QS)=PS^2-QS^2=PS^2-QS\cdot RS \), since \( QS=RS \).
3. Only \( PS^2 \) is not equal (it exceeds the product by \( QS^2 \)).

Teacher's Note:
1. Use the power of a point.
2. Write PQ and PR as PS minus and plus QS.