JEE Mathematics Circles MCQs Set 11

Answer: (a) none

Explanation:
1. \( S_1=-15<0 \), so the origin lies inside the circle, and no tangent can be drawn.

Teacher's Note:
1. No tangent can be drawn from an interior point.
2. S1 < 0 means inside.

Answer: (d) 0

Explanation:
1. The circle is \( (x+1)^2+(y-2)^2=1 \) with centre \( (-1,2) \).
2. The given point is the centre, which is inside: \( S_1=1+4-2-8+4=-1<0 \). No tangent can be drawn.

Teacher's Note:
1. Check if the point is the centre first.
2. An interior point has no tangents.

Answer: (a) 1

Explanation:
1. \( 1+4=5 \), so the point lies on the circle.
2. Exactly one tangent can be drawn from a point on the circle.

Teacher's Note:
1. S1 = 0 means on the circle.
2. A point on the circle has one tangent.

Answer: (d) one point is inside and the other is outside

Explanation:
1. \( S_1(0,0)=-11<0 \): inside.
2. \( S_1(1,8)=1+64-6+64-11=112>0 \): outside.

Teacher's Note:
1. Evaluate S1 for each point.
2. Negative = inside, positive = outside.

Answer: (c) \( a\in(-1,3) \)

Explanation:
1. The region is the upper half disc. Need \( y>0 \): \( a+1>0 \), so \( a>-1 \).
2. Inside the circle: \( a^2+(a+1)^2<25 \) gives \( a^2+a-12<0 \), so \( -4<a<3 \).
3. Both together: \( -1<a<3 \).

Teacher's Note:
1. The region is the upper semicircle, so two conditions are needed: above the x-axis and inside the circle.
2. Intersect the two intervals.

Answer: (d) \( (2\pi/3,4\pi/3) \)

Explanation:
1. Interior: \( (1+\cos\theta)^2+\sin^2\theta<1 \), i.e. \( 2+2\cos\theta<1 \).
2. \( \cos\theta<-\frac12 \), so \( \theta\in\left(\frac{2\pi}3,\frac{4\pi}3\right) \).

Teacher's Note:
1. Expand and use sin² + cos² = 1.
2. Solve the resulting trigonometric inequality on [0, 2π].

Answer: (d) empty set

Explanation:
1. Interior: \( 4+64-4+32-p<0 \), so \( p>96 \).
2. The centre is \( (1,-2) \), \( r^2=5+p \). Not meeting the axes needs \( r<1 \) (distance to the \( y\text{-axis} \)), i.e. \( 5+p<1 \), so \( p<-4 \).
3. No \( p \) can satisfy both \( p>96 \) and \( p<-4 \), so the set is empty.

Teacher's Note:
1. Find the condition for each requirement separately.
2. If the two ranges do not overlap the set is empty.

Answer: (c) \( r^2 \)

Explanation:
1. The tangent at \( (a,b) \) is \( ax+by=r^2 \), so \( \lambda=r^2 \).

Teacher's Note:
1. Tangent at (x1, y1) of x² + y² = r² is xx1 + yy1 = r².
2. Note that a² + b² = r² as well.

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

Explanation:
1. The radius to the origin has direction \( (a,b) \), so the tangent there is perpendicular to it: \( ax+by=0 \).

Teacher's Note:
1. The tangent is perpendicular to the radius at the point of contact.
2. The line through the origin with normal (a, b) is ax + by = 0.

Answer: (b) \( y \)-axis only

Explanation:
1. \( g=-4,\ f=2,\ c=4 \).
2. Touches the \( x\text{-axis} \) iff \( g^2=c \) (here \( 16\neq4 \)); touches the \( y\text{-axis} \) iff \( f^2=c \) (here \( 4=4 \)).

Teacher's Note:
1. Use g² = c for the x-axis and f² = c for the y-axis.
2. Check both conditions.

Answer: (c) \( p^2=a^2 \)

Explanation:
1. The distance from the origin to the line is \( |p| \) (since \( \cos^2\alpha+\sin^2\alpha=1 \)).
2. Tangency needs \( |p|=a \), i.e. \( p^2=a^2 \).

Teacher's Note:
1. The line is already in normal form, so p is its distance from the origin.
2. Tangency means distance = radius.

Answer: (b) \( y=mx\pm a\sqrt{1+m^2} \)

Explanation:
1. A parallel line \( y=mx+k \) is tangent when \( \frac{|k|}{\sqrt{1+m^2}}=a \).
2. So \( k=\pm a\sqrt{1+m^2} \).

Teacher's Note:
1. Tangent condition: distance from the origin equals the radius.
2. Both signs give the two parallel tangents.

Answer: (b) \( a^2(l^2+m^2)=n^2 \)

Explanation:
1. Distance of the centre from the line: \( \frac{|n|}{\sqrt{l^2+m^2}}=a \).
2. Squaring: \( n^2=a^2(l^2+m^2) \).

Teacher's Note:
1. Equate distance and radius, then square.
2. Do not mix the placement of n² and a².

Answer: (c) both the \( x \)-axis and \( y \)-axis

Explanation:
1. \( g=2,\ f=-2,\ c=4 \); \( g^2=4=c \) and \( f^2=4=c \).
2. Both conditions hold, so it touches both axes (centre \( (-2,2) \), radius 2).

Teacher's Note:
1. Check g² = c (x-axis) and f² = c (y-axis).
2. The centre (−2, 2) is at distance 2 from both axes.

Answer: (a) (1, -2)

Explanation:
1. The centre is \( (2,-3) \), \( r^2=4+9-11=2 \). The distance from the centre to the line is \( \frac{|2+3-3|}{\sqrt2}=\sqrt2 \), so it is a tangent.
2. The foot of the perpendicular from \( (2,-3) \): subtract \( \frac{(2)}{2}(1,-1) \) to get \( (1,-2) \).

Teacher's Note:
1. First confirm tangency with the distance, then find the foot of the perpendicular.
2. Direction of the normal is (1, −1).

Answer: (c) (0, 3)

Explanation:
1. Put \( x=0 \): \( y^2-6y+9=0 \), i.e. \( (y-3)^2=0 \).
2. A repeated root means tangency at \( (0,3) \).

Teacher's Note:
1. A tangent meets the circle in a repeated point.
2. Substitute x = 0 into the circle.

Answer: (d) \( \left(-\frac a{\sqrt2},\frac a{\sqrt2}\right) \)

Explanation:
1. The line is \( x-y+\sqrt2a=0 \), whose distance from the origin is \( a \), so it touches the circle.
2. The contact point is the foot of the perpendicular from the origin: \( -\frac{\sqrt2a}{2}(1,-1)=\left(-\frac a{\sqrt2},\frac a{\sqrt2}\right) \).

Teacher's Note:
1. The point of contact is the foot of the perpendicular from the centre.
2. Foot = −(c/(l² + m²)) (l, m) for the line lx + my + c = 0.

Answer: (b) \( 4y=3x \)

Explanation:
1. The radius direction at the origin is \( (-6,8) \), so the tangent is \( -6x+8y=0 \).
2. That is \( 3x=4y \), i.e. \( 4y=3x \).

Teacher's Note:
1. The tangent is perpendicular to the radius.
2. Simplify −6x + 8y = 0 by dividing by 2.

Answer: (a) (3, 0)

Explanation:
1. Tangents parallel to the \( y\text{-axis} \) are \( x=3 \) and \( x=-3 \).
2. \( x=-3 \) passes through the second and third quadrants, so the required tangent is \( x=3 \), touching at \( (3,0) \).

Teacher's Note:
1. Vertical tangents of x² + y² = 9 are x = ±3.
2. Choose the one that avoids the third quadrant.

Answer: (a) \( \left(-\frac52,\frac{5\sqrt3}2\right) \) and \( \left(\frac52,-\frac{5\sqrt3}2\right) \)

Explanation:
1. A tangent inclined at \( 30^\circ \) has direction \( (\cos30^\circ,\sin30^\circ) \), so the radius to the contact point is along the normal \( (-\sin30^\circ,\cos30^\circ)=\left(-\frac12,\frac{\sqrt3}2\right) \).
2. The contact points are \( \pm5\left(-\frac12,\frac{\sqrt3}2\right) \), i.e. \( \left(-\frac52,\frac{5\sqrt3}2\right) \) and \( \left(\frac52,-\frac{5\sqrt3}2\right) \).

Teacher's Note:
1. The radius at the contact point is perpendicular to the tangent.
2. Multiply the unit normal by the radius 5.

Answer: (c) -1

Explanation:
1. The radius to \( (h,h) \) has slope 1, so the tangent has slope \( -1 \) (valid for \( h\neq0 \)).

Teacher's Note:
1. Tangent slope is the negative reciprocal of the radius slope.
2. The radius slope is h/h = 1.

Answer: (c) \( (1,0) \) or \( \left(-\frac35,\frac45\right) \)

Explanation:
1. The chord of contact from \( (1,2) \) is \( x+2y=1 \).
2. Substituting \( x=1-2y \) in \( x^2+y^2=1 \): \( 5y^2-4y=0 \), so \( y=0 \) or \( \frac45 \).
3. The points are \( (1,0) \) and \( \left(-\frac35,\frac45\right) \).

Teacher's Note:
1. The two contact points lie on the chord of contact.
2. Intersect the chord of contact with the circle.

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

Explanation:
1. \( r=\frac{|2+3-1|}{\sqrt2}=2\sqrt2 \), so \( r^2=8 \).
2. Constant term \( =4+9-8=5 \): \( x^2+y^2-4x-6y+5=0 \).

Teacher's Note:
1. Compute r from the tangent distance.
2. Constant = h² + k² − r².

Answer: (b) (3, -1)

Explanation:
1. The tangent at \( (1,-2) \) is \( x-2y=5 \).
2. The second circle has centre \( (4,-3) \) and \( r^2=5 \); the distance from the centre to the line is \( \frac{|4+6-5|}{\sqrt5}=\sqrt5 \), so it touches.
3. The contact point is the foot of the perpendicular: \( (4,-3)-\frac55(1,-2)=(3,-1) \).

Teacher's Note:
1. Verify tangency by the distance, then find the foot of the perpendicular.
2. The normal direction is (1, −2).

Answer: (c) \( |c|=2 \)

Explanation:
1. The distance of the origin from \( y=c \) is \( |c| \), which must equal the radius 2.

Teacher's Note:
1. Tangency means distance = radius.
2. A horizontal line is at distance |c| from the origin.