JEE Mathematics Circles MCQs Set 08

Answer: (b) a circle

Explanation:
1. Take the square with sides \( x=0,\ x=1,\ y=0,\ y=1 \) and a moving point \( (x,y) \).
2. \( x^2+(x-1)^2+y^2+(y-1)^2=9 \) gives \( 2x^2+2y^2-2x-2y-7=0 \).
3. That is \( x^2+y^2-x-y-\frac72=0 \), a circle with centre \( (\frac12,\frac12) \) and radius \( 2 \).

Teacher's Note:
1. Sums of squared distances from fixed lines give a second-degree equation with equal coefficients of x² and y².
2. Such an equation always represents a circle if the radius is real.

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

Explanation:
1. The vertices are \( A(0,0),\ B(a,0),\ C(a,a),\ D(0,a) \).
2. The circle through them has diameter \( AC \), so its equation is \( x(x-a)+y(y-a)=0 \).
3. This is \( x^2+y^2-ax-ay=0 \).

Teacher's Note:
1. The circumcircle of a square has the diagonal as a diameter.
2. Use the diameter form of the circle with endpoints (0,0) and (a,a).

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

Explanation:
1. A circle touching the \( x \)-axis has radius equal to the \( y \)-coordinate of its centre, so \( r=2 \).
2. \( (x-1)^2+(y-2)^2=4 \) gives \( x^2+y^2-2x-4y+1=0 \).

Teacher's Note:
1. Touching the x-axis means the radius equals the distance of the centre from that axis.
2. Then use the centre-radius form.

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

Explanation:
1. \( r^2=(4-2)^2+(5-2)^2=13 \).
2. \( (x-2)^2+(y-2)^2=13 \) gives \( x^2+y^2-4x-4y+8-13=0 \), i.e. \( x^2+y^2-4x-4y-5=0 \).

Teacher's Note:
1. The radius is the distance between the centre and the point on the circle.
2. Do not forget the constant term: h² + k² − r².

Answer: (b) \( x^2+y^2-4x-10y+25=0 \)

Explanation:
1. Let the centre be \( (h,4h-3) \). Equal distances to the two points: \( (h-2)^2+(4h-6)^2=(h-4)^2+(4h-8)^2 \).
2. \( 17h^2-52h+40=17h^2-72h+80 \) gives \( h=2 \), centre \( (2,5) \).
3. \( r^2=(2-2)^2+(3-5)^2=4 \), so \( x^2+y^2-4x-10y+25=0 \).

Teacher's Note:
1. Impose the centre-on-line condition first to reduce to one unknown, then equate distances to the two points.
2. Check by substituting a given point in the final equation.

Answer: (c) \( a=b\neq0 \) and \( h=0 \)

Explanation:
1. A circle needs equal non-zero coefficients of x² and y² and no xy term.
2. So \( a=b\ne0 \) and \( h=0 \).

Teacher's Note:
1. Check the three features of a circle: equal squared-term coefficients, no xy term, real radius.
2. Options with a = b = 0 or only f = g do not ensure these.

Answer: (a) \( x^2+y^2-6x-5y+9=0 \)

Explanation:
1. The centre is \( (3,k) \) and the radius is \( |k| \).
2. \( (1-3)^2+(4-k)^2=k^2 \) gives \( 20-8k=0 \), \( k=\frac52 \).
3. \( (x-3)^2+(y-\frac52)^2=\frac{25}4 \) gives \( x^2+y^2-6x-5y+9=0 \).

Teacher's Note:
1. The centre lies on the vertical line through the point of contact.
2. Equate the distance to (1,4) with the radius to find k.

Answer: (a) only one circle

Explanation:
1. Three non-collinear points determine the circle through them uniquely, since the perpendicular bisectors of two chords meet at exactly one point (the centre).

Teacher's Note:
1. The general circle has three constants, so three conditions fix it.
2. If the points were collinear, no circle would pass through them.

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

Explanation:
1. The radius is the distance from \( (2,3) \) to \( x+y-1=0 \): \( r=\frac{|2+3-1|}{\sqrt2}=2\sqrt2 \).
2. \( r^2=8 \), so \( (x-2)^2+(y-3)^2=8 \).

Teacher's Note:
1. The radius equals the perpendicular distance from the centre to a tangent.
2. Square the distance for the equation, do not leave r on the right side.

Answer: (a) 3/4

Explanation:
1. The coefficients of \( (x+1)^2 \) and \( (y+2)^2 \) must be equal: \( \frac K3=\frac14 \).
2. So \( K=\frac34 \).

Teacher's Note:
1. A circle in centre form has equal coefficients for the two squared terms.
2. Then both terms combine to give r².

Answer: (a) \( x^2+y^2-8x-6y+16=0 \)

Explanation:
1. Let the centre be \( (h,h-1) \). Then \( (7-h)^2+(4-h)^2=9 \), i.e. \( h^2-11h+28=0 \), so \( h=4 \) or \( 7 \).
2. For \( h=4 \) the centre is \( (4,3) \): \( (x-4)^2+(y-3)^2=9 \) gives \( x^2+y^2-8x-6y+16=0 \).
3. (The other circle, centre \( (7,6) \), is not among the options.)

Teacher's Note:
1. Use the line to write the centre with one parameter and use the given point to get a quadratic for it.
2. Both roots give valid circles, so check which appears in the options.

Answer: (c) \( x^2+y^2+3x+12y+2=0 \)

Explanation:
1. Centre \( (h,2h-3) \): equal distances to the two points give \( 5h^2-10h+10=5h^2-8h+13 \), so \( h=-\frac32 \) and the centre is \( (-\frac32,-6) \).
2. So the circle is \( x^2+y^2+3x+12y+c=0 \); through \( (-2,0) \): \( 4-6+c=0 \), \( c=2 \).
3. Check with \( (3,-2) \): \( 9+4+9-24+2=0 \).

Teacher's Note:
1. Writing the centre as (h, 2h − 3) uses the line condition up front.
2. A final substitution check is quick and catches sign slips.

Answer: (b) \( x^2+y^2-10x-10y+25=0 \)

Explanation:
1. Centre \( (5,5) \) and radius \( 5 \).
2. \( (x-5)^2+(y-5)^2=25 \) gives \( x^2+y^2-10x-10y+25=0 \).

Teacher's Note:
1. Points of contact (5,0) and (0,5) give centre (5,5).
2. Radius equals 5.

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

Explanation:
1. \( r^2=\alpha^2+\beta^2 \).
2. \( (x-\alpha)^2+(y-\beta)^2=\alpha^2+\beta^2 \) gives \( x^2+y^2-2\alpha x-2\beta y=0 \).

Teacher's Note:
1. A circle through the origin has no constant term in the general form.
2. The linear coefficients are −2 times the centre coordinates.

Answer: (d) \( (x+5)^2+(y+5)^2=25 \)

Explanation:
1. In the third quadrant both coordinates of the centre are negative, and each has magnitude equal to the radius.
2. Centre \( (-5,-5) \): \( (x+5)^2+(y+5)^2=25 \).

Teacher's Note:
1. Centre signs follow the quadrant: third quadrant is (−, −).
2. The magnitude of each coordinate equals the radius.

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

Explanation:
1. Circle: \( x(x-a)+(y-1)(y-b)=0 \).
2. Put \( y=0 \): \( x^2-ax+b=0 \).

Teacher's Note:
1. Put y = 0 in the diameter form to find the x-axis intersections.
2. The constant term is (−1)(−b) = b.

Answer: (b) \( 0<k<1 \)

Explanation:
1. The circle through \( (0,0),(1,0),(0,1) \) is \( x^2+y^2-x-y=0 \).
2. Putting \( (2k,3k) \): \( 13k^2-5k=0 \), so \( k=0 \) or \( k=\frac5{13} \).
3. \( k=0 \) gives the origin again (not distinct), so only \( k=\frac5{13} \), which lies in \( 0<k<1 \).

Teacher's Note:
1. Find the circle through the three fixed points first, then test the fourth.
2. Exclude the value that makes points coincide.

Answer: (d) \( x^2+y^2+8x+4y+19=0 \)

Explanation:
1. The centre of the given circle is \( (-4,-2) \).
2. A unit circle with that centre: \( (x+4)^2+(y+2)^2=1 \), i.e. \( x^2+y^2+8x+4y+19=0 \).

Teacher's Note:
1. Concentric circles differ only in the constant term.
2. Constant = h² + k² − r² = 16 + 4 − 1 = 19.

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

Explanation:
1. The centre is \( (-r,r) \) with \( r\sqrt2=2\sqrt2 \), so \( r=2 \).
2. \( (x+2)^2+(y-2)^2=4 \) gives \( x^2+y^2+4x-4y+4=0 \).

Teacher's Note:
1. Distance of centre (r, r) from origin is r√2.
2. The quadrant fixes the signs of the centre coordinates.

Answer: (a) \( (x-1)^2+(y-1)^2=13 \)

Explanation:
1. Centre = midpoint \( (1,1) \).
2. \( r^2=(3-1)^2+(-2-1)^2=13 \), so \( (x-1)^2+(y-1)^2=13 \).

Teacher's Note:
1. The centre is the midpoint of the diameter.
2. Radius squared is the squared distance to either end.

Answer: (a) -11

Explanation:
1. Concentric circles share \( g=3,\ f=4 \).
2. Substitute \( (-3,2) \): \( 9+4-18+16+c=0 \), so \( c=-11 \).

Teacher's Note:
1. Concentric circles have identical g and f.
2. Use the given point on the circle to determine c.

Answer: (d) \( x^2+y^2-2x+2y-23=0 \)

Explanation:
1. The lines meet at \( (1,-1) \), the centre.
2. \( 2\pi r=10\pi \) gives \( r=5 \).
3. \( (x-1)^2+(y+1)^2=25 \) gives \( x^2+y^2-2x+2y-23=0 \).

Teacher's Note:
1. Diameters pass through the centre, so intersect them.
2. Circumference gives the radius.

Answer: (b) \( (1+m^2)(x^2+y^2)-2a(x+my)=0 \)

Explanation:
1. The given circle has centre \( (a,0) \): \( x^2+y^2-2ax=0 \).
2. Putting \( y=mx \): \( (1+m^2)x^2=2ax \), so the other end of the chord is \( \left(\frac{2a}{1+m^2},\frac{2am}{1+m^2}\right) \).
3. Diameter form: \( x^2+y^2-\frac{2a}{1+m^2}(x+my)=0 \), i.e. \( (1+m^2)(x^2+y^2)-2a(x+my)=0 \).

Teacher's Note:
1. Find the second end of the chord by intersecting the line with the circle.
2. Then use the diameter form with the origin as the first end.

Answer: (a) \( x^2+y^2-a^2-2p(x\cos\alpha+y\sin\alpha-p)=0 \)

Explanation:
1. Circles through the intersections of the circle and the line: \( x^2+y^2-a^2+\lambda(x\cos\alpha+y\sin\alpha-p)=0 \).
2. Its centre \( \left(-\frac{\lambda\cos\alpha}{2},-\frac{\lambda\sin\alpha}{2}\right) \) must lie on the chord (a diameter of the new circle lies along the chord): \( -\frac\lambda2-p=0 \), so \( \lambda=-2p \).
3. This gives \( x^2+y^2-a^2-2p(x\cos\alpha+y\sin\alpha-p)=0 \).

Teacher's Note:
1. The chord is a diameter of the new circle, so the centre must lie on the chord.
2. Use the family S + λL = 0 and fix λ with this condition.

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

Explanation:
1. The centre is \( (-r,-r) \) and the radius is \( r \).
2. Distance to the line: \( \frac{|-3r+4r+8|}{5}=r \), so \( |r+8|=5r \), giving \( r=2 \) (the other sign gives a negative radius).
3. \( (x+2)^2+(y+2)^2=4 \) gives \( x^2+y^2+4x+4y+4=0 \).

Teacher's Note:
1. Write the centre using the quadrant, then apply the tangent-distance condition.
2. Reject any negative radius.