JEE Mathematics Circles MCQs Set 10

Answer: (a) \( x=1,\ y=4 \)

Explanation:
1. The centre is the midpoint: \( \frac{x+3}2=2 \) gives \( x=1 \) and \( \frac{3+5}2=y \) gives \( y=4 \).

Teacher's Note:
1. Equate the midpoint coordinates to the centre.
2. Solve the two simple equations separately.

Answer: (c) (1, 2)

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

Teacher's Note:
1. The centre is the midpoint of any diameter.
2. Reflect the given end in the centre.

Answer: (b) \( \frac{17}{6\sqrt5} \)

Explanation:
1. The lines are parallel: \( 2x-4y-9=0 \) and \( 2x-4y+\frac73=0 \).
2. Distance between them \( =\frac{|-9-\frac73|}{\sqrt{20}}=\frac{34/3}{2\sqrt5}=\frac{17}{3\sqrt5} \).
3. This is the diameter, so the radius is \( \frac{17}{6\sqrt5} \).

Teacher's Note:
1. Two parallel tangents are a diameter apart.
2. Make the x and y coefficients equal before using the distance formula.

Answer: (d) None of these

Explanation:
1. No \( xy \) term means \( \lambda=0 \), giving \( 2x^2+2y^2-4x+6y-5=0 \), i.e. \( x^2+y^2-2x+3y-\frac52=0 \).
2. \( r^2=1+\frac94+\frac52=\frac{23}4 \), so \( r=\frac{\sqrt{23}}2 \).
3. This is none of the listed values.

Teacher's Note:
1. The xy coefficient must vanish for a circle.
2. Then compute r from g, f, c.

Answer: (a) \( 2\sqrt{11} \)

Explanation:
1. The circle has centre \( (5,7) \) and radius \( \sqrt{25+49+51}=5\sqrt5 \). The distance from \( (-7,2) \) to the centre is \( \sqrt{144+25}=13 \).
2. \( 13>5\sqrt5\ (\approx11.2) \), so the point is outside: \( p=13+5\sqrt5,\ q=13-5\sqrt5 \).
3. \( pq=169-125=44 \), so GM \( =\sqrt{44}=2\sqrt{11} \).

Teacher's Note:
1. The product pq equals the power of the point, d² − r².
2. Check first whether the point is outside the circle.

Answer: (a) \( \left(\frac{7}{2\sqrt2},\frac7{2\sqrt2}\right) \)

Explanation:
1. The distance from the origin to the line is \( \frac{5\sqrt2}{\sqrt2}=5 \); the given circle has radius 2.
2. The gap between the circle and the line is \( 5-2=3 \), so the smallest touching circle has diameter 3 and radius \( \frac32 \).
3. Its centre lies on the perpendicular from the origin at distance \( 2+\frac32=\frac72 \): \( \frac72\cdot\frac{(1,1)}{\sqrt2}=\left(\frac7{2\sqrt2},\frac7{2\sqrt2}\right) \).

Teacher's Note:
1. The smallest circle fits exactly in the gap along the common perpendicular.
2. The centre lies along the direction (1, 1) from the origin.

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

Explanation:
1. Put \( u=x\cos\theta+y\sin\theta \) and \( v=x\sin\theta-y\cos\theta \); this is a rotation, so \( u^2+v^2=x^2+y^2 \).
2. The circle is \( (u-a)^2+(v+a)^2=a^2 \), with centre \( u=a,\ v=-a \).
3. For the centre \( x^2+y^2=u^2+v^2=2a^2 \), whatever \( \theta \) is.

Teacher's Note:
1. The substitution u, v is a rotation of axes, which preserves distance from the origin.
2. The centre is at distance a√2 from the origin for every θ.

Answer: (c) 5

Explanation:
1. The centre is \( (r,4) \) and its distance from the \( x \)-axis is 4.
2. \( r^2=4^2+3^2=25 \), so \( r=5 \).

Teacher's Note:
1. Half the chord and the distance from the centre to the chord form a right triangle with the radius.
2. Touching the y-axis at (0, 4) fixes the centre's y-coordinate as 4.

Answer: (c) 5

Explanation:
1. The centre is \( (r,3) \); its distance from the \( x \)-axis is 3.
2. \( r^2=3^2+4^2=25 \), so \( r=5 \).

Teacher's Note:
1. Use half the intercept (4) as one leg of a right triangle.
2. The centre height is the y-coordinate of the contact point.

Answer: (d) (1, 0), (2, 0)

Explanation:
1. Put \( y=0 \): \( x^2-3x+2=0 \), so \( x=1 \) or \( 2 \).

Teacher's Note:
1. Set y = 0 to find the x-axis intersections.
2. Factorise the quadratic.

Answer: (b) \( \pm4 \)

Explanation:
1. A circle touches the \( x \)-axis iff \( g^2=c \).
2. \( a^2=16 \), so \( a=\pm4 \).

Teacher's Note:
1. Setting y = 0 gives x² + 2ax + 16 = 0, which needs a zero discriminant.
2. That is a² = 16.

Answer: (b) \( \sqrt2 \)

Explanation:
1. The distance of the centre from the line is \( \frac1{\sqrt2} \).
2. Chord \( =2\sqrt{1-\frac12}=\sqrt2 \).

Teacher's Note:
1. Chord length = 2√(r² − d²).
2. Compute d from the centre to the line.

Answer: (a) no points

Explanation:
1. The line is \( x+y+1=0 \). The distance from the centre \( (2,3) \) is \( \frac{6}{\sqrt2}=3\sqrt2\approx4.24 \).
2. The radius is \( \sqrt{11}\approx3.32<4.24 \), so the line does not meet the circle.

Teacher's Note:
1. Compare the distance from the centre to the line with the radius.
2. Distance greater than radius means no intersection.

Answer: (a) (-2, -6), (4, 2)

Explanation:
1. \( y=\frac{4x-10}3 \); substituting and multiplying by 9: \( 25x^2-50x-200=0 \), i.e. \( x^2-2x-8=0 \).
2. \( x=4 \) gives \( y=2 \); \( x=-2 \) gives \( y=-6 \).

Teacher's Note:
1. Substitute y from the line into the circle.
2. Check both points in the line equation.

Answer: (c) both (a) and (b) together, i.e. the full interval

Explanation:
1. Distance from the origin to the line: \( \frac{|c|}{\sqrt{1+m^2}}<r \), so \( |c|<r\sqrt{1+m^2} \).
2. That is \( -r\sqrt{1+m^2}<c<r\sqrt{1+m^2} \), the union of the intervals in (a) and (b).

Teacher's Note:
1. Two distinct points need the distance from the centre to be less than the radius.
2. The modulus condition splits into negative and non-negative c.

Answer: (c) infinite

Explanation:
1. The line passes through \( (1,1) \), and \( 1^2+1^2=2<4 \), so \( (1,1) \) lies inside the circle.
2. Every line through an interior point cuts the circle at two real points, for every value of \( m \).

Teacher's Note:
1. A line through a point inside a circle always meets it twice.
2. Check the point using x² + y² compared with r².

Answer: (a) \( ax+by=0 \)

Explanation:
1. The circle passes through \( (0,0),(2a,0),(0,2b) \): \( x^2+y^2-2ax-2by=0 \).
2. The tangent at the origin is \( -ax-by=0 \), i.e. \( ax+by=0 \).

Teacher's Note:
1. The tangent at the origin of x² + y² + 2gx + 2fy = 0 is gx + fy = 0.
2. Here g = −a and f = −b.

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

Explanation:
1. It passes through \( (0,0),(a,0),(0,b) \).
2. With the angle at the origin being a right angle, the diameter joins \( (a,0) \) and \( (0,b) \): \( x^2+y^2-ax-by=0 \).

Teacher's Note:
1. Intercepts are the points where the circle meets the axes.
2. Use the right angle at the origin.

Answer: (b) \( y=3,\ 12x+5y=39 \)

Explanation:
1. A chord of length 8 is at distance \( \sqrt{25-16}=3 \) from the origin.
2. \( y=3 \) is at distance 3 and passes through \( (2,3) \). For \( y-3=m(x-2) \): \( \frac{|3-2m|}{\sqrt{1+m^2}}=3 \) gives \( 5m^2+12m=0 \), so \( m=0 \) or \( -\frac{12}5 \).
3. The second line is \( 12x+5y=39 \).

Teacher's Note:
1. First convert the chord length into the distance from the centre.
2. Then impose that distance on a line through the given point.

Answer: (a) \( x^2+y^2-4(\sqrt2x+y)+4=0 \)

Explanation:
1. The centre is \( (h,2) \) with radius \( h \) (touching the \( y \)-axis).
2. Intercept 4: \( h^2=2^2+2^2=8 \), so \( h=2\sqrt2 \) (positive side).
3. \( (x-2\sqrt2)^2+(y-2)^2=8 \) gives \( x^2+y^2-4\sqrt2x-4y+4=0 \).

Teacher's Note:
1. Half the intercept and the centre height give the radius.
2. The radius also equals the horizontal distance from the y-axis.

Answer: (c) the y-axis

Explanation:
1. The intercept lies from \( x=-3 \) to \( x=3 \), so the circle passes through \( (3,0) \) and \( (-3,0) \).
2. Its centre is on the perpendicular bisector of these points, the \( y \)-axis.

Teacher's Note:
1. The two points where the circle meets the x-axis are (3, 0) and (−3, 0).
2. The centre lies on their perpendicular bisector.

Answer: (b) \( (1+m^2)(a^2-l^2) \)

Explanation:
1. Distance from the origin to the line: \( d^2=\frac{c^2}{1+m^2} \).
2. \( l^2=a^2-d^2 \), so \( c^2=(1+m^2)(a^2-l^2) \).

Teacher's Note:
1. Half the chord is l, so l² = a² − d².
2. Solve for c².

Answer: (b) 2

Explanation:
1. The line is \( 4x+3y-12=0 \); its distance from the origin is \( \frac{12}5 \).
2. \( r^2=\frac{169}{25} \), so chord \( =2\sqrt{\frac{169-144}{25}}=2 \).

Teacher's Note:
1. Chord = 2√(r² − d²).
2. Get d from the intercept form.

Answer: (b) \( \alpha^2+\beta^2-r^2 \)

Explanation:
1. The product of the lengths of the two segments of any secant from an outside point equals the square of the tangent length.
2. That is the power of the point: \( \alpha^2+\beta^2-r^2 \).

Teacher's Note:
1. Power of a point is constant for all secants through it.
2. It equals S1 = α² + β² − r².

Answer: (b) (-2, 1)

Explanation:
1. A point is inside iff \( S_1<0 \). For \( (-2,1) \): \( 4+1-6-3+2=-2<0 \).
2. The others give \( S_1=0,\ 14,\ 0 \) respectively (\( (-1,3) \) and \( (-3,2) \) lie on the circle, \( (2,1) \) is outside).

Teacher's Note:
1. Substitute each point into x² + y² + 3x − 3y + 2.
2. Negative means inside, zero means on the circle.