JEE Mathematics Circles MCQs Set 09

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

Explanation:
1. Each option has radius 2 and centre \( (\pm2,\pm2) \); test the distance of each centre from \( 3x+4y-4=0 \) (it must equal 2).
2. Centre \( (2,2) \): \( \frac{|6+8-4|}{5}=2 \), which works. The centres \( (2,-2),\ (-2,-2),\ (-2,2) \) give distances \( \frac65,\ \frac{18}5,\ \frac25 \), none equal to 2.
3. So the circle is \( (x-2)^2+(y-2)^2=4 \), i.e. \( x^2+y^2-4x-4y+4=0 \).

Teacher's Note:
1. A circle tangent to a line has its centre at distance r from it.
2. Testing the four candidate centres is quicker than solving afresh.

Answer: (d) \( (x-p)^2=4qy \)

Explanation:
1. Let the other end be \( (h,k) \). The centre is \( \left(\frac{p+h}2,\frac{q+k}2\right) \) and, since the circle touches the \( x \)-axis, the radius is \( \frac{q+k}2 \).
2. The radius also equals half the diameter: \( \frac{(p-h)^2+(q-k)^2}{4}=\frac{(q+k)^2}{4} \).
3. This gives \( (h-p)^2=4qk \), so the locus is \( (x-p)^2=4qy \).

Teacher's Note:
1. Express the radius two ways: as the y-coordinate of the centre and as half of AB.
2. The locus is a parabola.

Answer: (c) Both (a) and (b)

Explanation:
1. The lines are perpendicular, so the centre is at distance \( r \) from both. Check option (a): centre \( (2,8) \), \( r=5 \). Distances: \( \frac{|6-32+1|}{5}=5 \) and \( \frac{|8+24-7|}{5}=5 \); and \( (2,3) \) is at distance 5. Valid.
2. Option (b): centre \( (\frac65,\frac{12}5) \), \( r=1 \). Distances: \( \frac{|3.6-9.6+1|}5=1 \) and \( \frac{|4.8+7.2-7|}5=1 \); and substituting \( (2,3) \): \( 20+45-24-72+31=0 \). Valid.
3. Both circles satisfy all conditions.

Teacher's Note:
1. Two perpendicular tangent lines give two circles through a point in the same angular region.
2. Verify each option against all three conditions.

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

Explanation:
1. Any circle through two points has radius at least half the distance between them, with equality when the segment is a diameter.
2. The diameter form with ends \( (1,0),(0,1) \): \( (x-1)x+y(y-1)=0 \), i.e. \( x^2+y^2-x-y=0 \).

Teacher's Note:
1. A chord is never longer than the diameter.
2. So the smallest circle has the given chord as diameter.

Answer: (a) \( 4x^2+4y^2-4x-4y+1=0 \)

Explanation:
1. The centre is \( (r,r) \); being below the line, \( 4r+3r<6 \), so \( 6-7r>0 \).
2. Distance to the line: \( \frac{6-7r}{5}=r \), so \( r=\frac12 \).
3. \( (x-\frac12)^2+(y-\frac12)^2=\frac14 \) gives \( 4x^2+4y^2-4x-4y+1=0 \).

Teacher's Note:
1. Being below the line fixes the sign inside the modulus.
2. Then 6 − 7r = 5r.

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

Explanation:
1. \( x^2-y^2-2x+4y-3=(x-1)^2-(y-2)^2 \), so the lines are \( y-2=\pm(x-1) \) and meet at the centre \( (1,2) \).
2. \( r^2=(1-1)^2+(1-2)^2=1 \).
3. \( (x-1)^2+(y-2)^2=1 \) gives \( x^2+y^2-2x-4y+4=0 \).

Teacher's Note:
1. Complete the squares to factor the pair of lines and find their intersection.
2. The point of intersection of two diameters is the centre.

Answer: (a) \( 9(x^2+y^2)+6x+24y+1=0 \)

Explanation:
1. The centre is \( (h,h-1) \) with \( h<0 \) and radius \( |h-1|=1-h \).
2. Distance to the line: \( \frac{|4h-3(h-1)+4|}{5}=\frac{|h+7|}5=1-h \), so \( h+7=5-5h \) and \( h=-\frac13 \) (the other sign gives \( h=3 \), not in the third quadrant).
3. Centre \( (-\frac13,-\frac43) \), \( r=\frac43 \): \( 9(x^2+y^2)+6x+24y+1=0 \).

Teacher's Note:
1. The quadrant fixes the sign of the radius expression.
2. Solve the linear equation for h and reject the root outside the quadrant.

Answer: (b) \( x^2+y^2-3x-5y+8=0 \)

Explanation:
1. \( xy-2x-y+2=(x-1)(y-2) \), so the lines are \( x+y=4,\ x=1,\ y=2 \).
2. The vertices are \( (1,2),\ (1,3),\ (2,2) \); the angle at \( (1,2) \) is a right angle, so \( (1,3)(2,2) \) is a diameter.
3. \( (x-1)(x-2)+(y-3)(y-2)=0 \) gives \( x^2+y^2-3x-5y+8=0 \).

Teacher's Note:
1. Factorise the product to find the three lines.
2. A right-angled triangle has its hypotenuse as the diameter of the circumcircle.

Answer: (c) \( x^2+y^2+4x+2y-15=0 \)

Explanation:
1. If \( P(x_1,y_1),Q(x_2,y_2) \), then \( x_1+x_2=-4,\ x_1x_2=-12 \) and \( y_1+y_2=-2,\ y_1y_2=-3 \).
2. Diameter form: \( x^2-(x_1+x_2)x+x_1x_2+y^2-(y_1+y_2)y+y_1y_2=0 \).
3. \( x^2+4x-12+y^2+2y-3=0 \), i.e. \( x^2+y^2+4x+2y-15=0 \).

Teacher's Note:
1. The diameter form depends only on the sums and products of the abscissae and ordinates, so the pairing of roots does not matter.
2. Read the roots as sums and products from the quadratics.

Answer: (b) \( 4(x^2+y^2)+8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f) \)

Explanation:
1. The incircle is concentric with the circumcircle (centre \( (-g,-f) \)) and has radius \( \frac R2 \), where \( R^2=(1+g)^2+(1+f)^2 \) (distance from the centre to the vertex \( (1,1) \)).
2. Incircle: \( (x+g)^2+(y+f)^2=\frac{R^2}4 \), i.e. \( 4(x^2+y^2)+8gx+8fy=(1+g)^2+(1+f)^2-4g^2-4f^2 \).
3. Since \( (1+g)^2-4g^2=(1-g)(1+3g) \) (and similarly for \( f \)), this is option (b).

Teacher's Note:
1. In an equilateral triangle r = R/2 and both circles share the centre.
2. Use the vertex to get R, not the constant c.

Answer: (a) \( x^2+y^2-3ax+(1-\sqrt2)ay+(2-\sqrt2)a^2=0 \)

Explanation:
1. The quadrilateral is a rectangle with vertices \( (a,-a),(2a,-a),(2a,\sqrt2a),(a,\sqrt2a) \); its circumcircle has the diagonal from \( (a,-a) \) to \( (2a,\sqrt2a) \) as a diameter.
2. \( (x-a)(x-2a)+(y+a)(y-\sqrt2a)=0 \).
3. Expanding: \( x^2+y^2-3ax+(1-\sqrt2)ay+(2-\sqrt2)a^2=0 \).

Teacher's Note:
1. The four lines form a rectangle, whose circumcircle has a diagonal as diameter.
2. Expand the diameter form carefully.

Answer: (c) \( 25\pi \)

Explanation:
1. \( r^2=(4-1)^2+(6-2)^2=25 \), so \( r=5 \).
2. Area \( =\pi r^2=25\pi \).

Teacher's Note:
1. The radius is the distance from the centre to the point.
2. Area is πr², so only r² is needed.

Answer: (b) collinear

Explanation:
1. The centres are \( (0,0),\ (-3,1) \) and \( (6,-2) \).
2. Slope from \( (0,0) \) to \( (-3,1) \) is \( -\frac13 \); slope from \( (0,0) \) to \( (6,-2) \) is \( -\frac13 \).
3. Equal slopes, so the three centres lie on one line.

Teacher's Note:
1. Read the centre as (−g, −f) for each circle.
2. Equal slopes from a common point imply collinearity.

Answer: (c) -4

Explanation:
1. \( g=-9,\ f=6 \), so \( r^2=81+36-k=117-k \).
2. \( 117-k=121 \) gives \( k=-4 \).

Teacher's Note:
1. Use r² = g² + f² − c.
2. Set it equal to 121.

Answer: (a) (3, 4)

Explanation:
1. Compare with \( (x-h)^2+(y-k)^2=r^2 \): \( h=3,\ k=4 \).

Teacher's Note:
1. The centre is (h, k), not (−h, −k).
2. The right side gives r² = 5.

Answer: (a) Centre (-1, -2), radius = 2

Explanation:
1. \( g=1,\ f=2,\ c=1 \), so the centre is \( (-1,-2) \).
2. \( r=\sqrt{1+4-1}=2 \).

Teacher's Note:
1. Centre is (−g, −f).
2. Radius is √(g² + f² − c).

Answer: (b) 3

Explanation:
1. \( g=\cos\theta,\ f=\sin\theta \), so \( r^2=\cos^2\theta+\sin^2\theta+8=9 \).
2. \( r=3 \).

Teacher's Note:
1. Use r² = g² + f² − c.
2. cos² + sin² = 1.

Answer: (d) (11, 2)

Explanation:
1. The centre is \( (4,2) \).
2. Other end \( =2(4,2)-(-3,2)=(11,2) \).

Teacher's Note:
1. The centre is the midpoint of the diameter.
2. Other end = 2 × centre − given end.

Answer: (b) (-1, 3)

Explanation:
1. \( (x+1)^2+(y-3)^2=4(\cos^2\theta+\sin^2\theta)=4 \).
2. The centre is \( (-1,3) \).

Teacher's Note:
1. Eliminate θ using cos² + sin² = 1.
2. In parametric form (h + r cos θ, k + r sin θ) the centre is (h, k).

Answer: (a) (4, 7)

Explanation:
1. \( x^2-8x+12=0 \) gives \( x=2,6 \) and \( y^2-14y+45=0 \) gives \( y=5,9 \).
2. The square is \( [2,6]\times[5,9] \) (side 4), so the centre is \( (4,7) \).

Teacher's Note:
1. Factorise each quadratic to get the four sides.
2. The incircle of a square has the same centre as the square.

Answer: (b) \( x+3y=0 \)

Explanation:
1. The centre is \( (6,-2) \).
2. Check \( x+3y=6-6=0 \): the centre lies on this line, so it is a diameter. The other lines give \( 4,\ 8,\ 14 \), not 0.

Teacher's Note:
1. A diameter is any line through the centre.
2. Substitute the centre in each option.

Answer: (b) \( 2\sqrt5 \)

Explanation:
1. Solving \( x+y=6,\ x+2y=4 \): \( y=-2,\ x=8 \); centre \( (8,-2) \).
2. \( r^2=(8-6)^2+(-2-2)^2=20 \), so \( r=2\sqrt5 \).

Teacher's Note:
1. The intersection of two diameters is the centre.
2. Use the distance from the centre to the given point.

Answer: (d) None of these

Explanation:
1. Divide by 2: \( x^2+y^2+2\lambda x+\frac{\lambda^2}2=0 \).
2. \( r^2=\lambda^2-\frac{\lambda^2}2=\frac{\lambda^2}2>0 \) exactly when \( \lambda\neq0 \).
3. So the range is all real \( \lambda \) except 0, which is none of (a), (b), (c).

Teacher's Note:
1. Compute g² − c and demand it be positive.
2. λ = 0 gives a point circle, so it must be excluded.

Answer: (a) \( A=(2-2\sqrt2,-3-2\sqrt2),\ B=(2+2\sqrt2,-3+2\sqrt2) \)

Explanation:
1. The circle has centre \( (2,-3) \) and radius \( 4 \). The point \( (7,2) \) lies along direction \( (1,1) \) from the centre.
2. The farthest and nearest points lie on the line through the centre in direction \( (1,1) \): centre \( \pm4\cdot\frac{(1,1)}{\sqrt2} \).
3. Nearest: \( (2+2\sqrt2,-3+2\sqrt2) \); farthest: \( (2-2\sqrt2,-3-2\sqrt2) \).

Teacher's Note:
1. The nearest and farthest points of a circle from an outside point lie on the line joining that point and the centre.
2. Move 4 units from the centre along that direction (nearest) or opposite to it (farthest).

Answer: (b) \( \sqrt5 \)

Explanation:
1. The centre is \( (4,6) \).
2. \( r^2=(5-4)^2+(4-6)^2=5 \), so \( r=\sqrt5 \).

Teacher's Note:
1. Concentric circles share the centre.
2. The radius is the distance from the centre to the given point.