JEE Mathematics Circle MCQs Set 01

Answer: (d) none of these

Explanation:
1. There is no xy term in a circle, so \( \lambda=0 \).
2. Then \( x^2+y^2-2x+3y-\frac52=0 \) with centre \( \left(1,-\frac32\right) \).
3. \( r^2=1+\frac94+\frac52=\frac{23}4 \), so \( r=\frac{\sqrt{23}}2\), which is none of the listed values.

Teacher's Note:
1. The xy coefficient must vanish.
2. Use r squared = g squared + f squared - c.

Answer: (a) a point

Explanation:
1. \( x^2-2x+1+y^2+4y+4=0\Rightarrow(x-1)^2+(y+2)^2=0 \).
2. Only the point \( (1,-2) \) satisfies it.

Teacher's Note:
1. Complete the squares.
2. A sum of squares equal to zero gives a single point.

Answer: (c) both (a) and (b) hold together

Explanation:
1. The curve passes through the three vertices. It is a circle when the \( xy \) coefficient is zero and the coefficients of \( x^2 \) and \( y^2 \) are equal.
2. The \( xy \) coefficient gives \( \sum\lambda(m_2+m_3)=0 \); equality of the \( x^2 \) and \( y^2 \) coefficients gives \( \sum\lambda(m_2m_3-1)=0 \).
3. Both conditions are needed.

Teacher's Note:
1. A circle has no xy term and equal squared coefficients.
2. The curve through the vertices already passes through the three points.

Answer: (d) none of these

Explanation:
1. \( r^2=\frac{\lambda^2}{4}+\frac{(1-\lambda)^2}{4}-5=\frac{2\lambda^2-2\lambda+1}{4}-5 \). We need \( 0\lt r^2\le25 \).
2. \( r^2>0\Rightarrow\lambda^2-\lambda-9.5>0\Rightarrow\lambda\lt -2.62 \) or \( \lambda>3.62 \). \( r^2\le25\Rightarrow\lambda^2-\lambda-59.5\le0\Rightarrow-7.2\le\lambda\le8.2 \).
3. Integers: \( -7,-6,-5,-4,-3 \) and \( 4,5,6,7,8 \), i.e. 10 values, which is not listed.

Teacher's Note:
1. The expression must be positive for a real circle.
2. Do not forget the lower bound on r squared.

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 \( x-1=\pm(y-2) \), meeting at the centre \( (1,2) \).
2. Radius \( =\) distance to \( (1,1)=1 \).
3. \( (x-1)^2+(y-2)^2=1\Rightarrow x^2+y^2-2x-4y+4=0 \).

Teacher's Note:
1. Diameters meet at the centre.
2. Factor the pair of lines as a difference of squares.

Answer: (c) \( \sqrt3(x^2+y^2)-2y-\sqrt3=0 \)

Explanation:
1. The third vertex is \( (0,\sqrt3) \); the centroid (also the circumcentre) is \( \left(0,\frac1{\sqrt3}\right) \) with \( R=\frac2{\sqrt3} \).
2. \( x^2+\left(y-\frac1{\sqrt3}\right)^2=\frac43\Rightarrow x^2+y^2-\frac2{\sqrt3}y-1=0 \).
3. Multiplying by \( \sqrt3 \): \( \sqrt3(x^2+y^2)-2y-\sqrt3=0 \).

Teacher's Note:
1. Locate the third vertex first.
2. The circumcentre of an equilateral triangle is its centroid.

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. Vertices: \( (1,3),\ (2,2),\ (1,2) \). The triangle is right angled at \( (1,2) \), so the hypotenuse from \( (1,3) \) to \( (2,2) \) is a diameter.
3. \( (x-1)(x-2)+(y-3)(y-2)=0\Rightarrow x^2+y^2-3x-5y+8=0 \).

Teacher's Note:
1. Factor the second bracket.
2. A right angle in a triangle means the hypotenuse is the circumdiameter.

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

Explanation:
1. Abscissae are \( -6 \) and \( 2 \); ordinates are \( -3 \) and \( 1 \). Taking \( P=(-6,-3) \) and \( Q=(2,1) \) (the other pairing gives the same circle).
2. Centre \( =(-2,-1) \), \( r^2=\frac{64+16}{4}=20 \).
3. \( (x+2)^2+(y+1)^2=20\Rightarrow x^2+y^2+4x+2y-15=0 \).

Teacher's Note:
1. Find the coordinate values from each quadratic.
2. Both possible pairings give the same centre and the same diameter length.

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

Explanation:
1. Touching the y-axis at \( (0,2) \): centre \( (r,2) \). The x-axis chord is \( 2\sqrt{r^2-4}=4\Rightarrow r^2=8,\ r=2\sqrt2 \).
2. \( (x-2\sqrt2)^2+(y-2)^2=8\Rightarrow x^2+y^2-4\sqrt2x-4y+4=0 \).

Teacher's Note:
1. The point of contact fixes the y-coordinate of the centre.
2. Use the chord length formula for the x-axis.

Answer: (b) \( 3+2\sqrt2 \)

Explanation:
1. The centres are \( (1,1) \) and \( (r,r) \). They touch externally: \( \sqrt2(r-1)=r+1 \).
2. \( r(\sqrt2-1)=\sqrt2+1\Rightarrow r=\frac{\sqrt2+1}{\sqrt2-1}=(\sqrt2+1)^2=3+2\sqrt2 \).

Teacher's Note:
1. A circle touching both axes in the first quadrant has centre (r, r).
2. Distance between centres equals the sum of the radii.

Answer: (c) (4, 5), (-2, -3)

Explanation:
1. Both centres lie on the normal to the tangent at \( (1,1) \), at distance 5 from it.
2. Unit normal \( \frac{(3,4)}5 \). Centres: \( (1,1)\pm5\cdot\frac{(3,4)}{5}=(4,5)\) and \( (-2,-3) \).

Teacher's Note:
1. The radius to the point of contact is perpendicular to the tangent.
2. Move 5 units along the unit normal in both directions.

Answer: (a) \( x^2+y^2-2\sqrt3y-1=0 \)

Explanation:
1. The side is 2, so the circumradius is 2. The centre lies on the y-axis at height \( \sqrt{4-1}=\sqrt3 \).
2. \( x^2+(y-\sqrt3)^2=4\Rightarrow x^2+y^2-2\sqrt3y-1=0 \).

Teacher's Note:
1. In a regular hexagon the circumradius equals the side.
2. The centre is on the perpendicular bisector of the given side.

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

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

Teacher's Note:
1. Longest and shortest distances are d plus r and d minus r (in absolute value).
2. The product is d squared minus r squared in magnitude.

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, with radius \( \frac R2 \). The vertex \( (1,1) \) is on the circumcircle: \( c=-2-2g-2f \), so \( R^2=(1+g)^2+(1+f)^2 \).
2. Incircle: \( x^2+y^2+2gx+2fy+g^2+f^2-\frac{R^2}4=0 \).
3. Multiplying by 4 and simplifying gives \( 4(x^2+y^2)+8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f) \).

Teacher's Note:
1. In an equilateral triangle r = R/2 and the centres coincide.
2. Use the vertex to find c.

Answer: (b) \( \left(\frac{29}{10},2\right) \)

Explanation:
1. Equal roots on the y-axis means the circle touches the y-axis at \( (0,2) \), so the centre has \( y=2 \).
2. The x-intercepts \( \frac45 \) and \( 5 \) are symmetric about the centre: \( h=\frac{\frac45+5}2=\frac{29}{10} \).
3. Centre \( =\left(\frac{29}{10},2\right) \).

Teacher's Note:
1. Equal roots mean tangency to that axis.
2. The centre lies midway between the intercepts.

Answer: (a) 7

Explanation:
1. Moving k units on a circle of radius k turns through 1 radian each time.
2. After the circle \( C_n \), the total angle is \( n \) radians. It first exceeds \( 2\pi\approx6.28 \) when \( n=7 \).

Teacher's Note:
1. Arc length divided by radius gives the angle in radians.
2. Add the angles on successive circles.

Answer: (a) 1

Explanation:
1. On the y-axis \( x=0 \): \( y^2-7y+12=0\Rightarrow y=3,4 \).
2. The chord length is \( 4-3=1 \).

Teacher's Note:
1. Put x equal to zero to get the points on the y-axis.
2. Subtract the ordinates.

Answer: (c) infinite

Explanation:
1. The point \( (1,1) \) is inside the circle since \( 1+1\lt 4 \).
2. Every line through an interior point cuts the circle at two real points, for every real m.
3. So there are infinitely many values of m.

Teacher's Note:
1. Check whether the fixed point lies inside the circle.
2. Any line through an interior point is a secant.

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

Explanation:
1. The chord has slope 1: \( y=x-3 \). Its distance from the origin is \( \frac3{\sqrt2} \).
2. Length \( =2\sqrt{9-\frac92}=2\cdot\frac3{\sqrt2}=3\sqrt2 \).

Teacher's Note:
1. Find the line first, then its distance from the centre.
2. Use chord length = 2 root of (r squared - d squared).

Answer: (b) 0

Explanation:
1. The circle has centre \( \left(\frac34,0\right) \) and radius \( \frac34 \). The distance from \( (-2,1) \) to the centre is \( \sqrt{\frac{121}{16}+1}=\sqrt{\frac{137}{16}}\approx2.93 \).
2. The points at distance 2 form a circle of radius 2 about \( (-2,1) \). The circles meet only if \( 2.93\le2+0.75=2.75 \), which is false.
3. So there are no such points.

Teacher's Note:
1. Compare the centre distance with the sum of the radii.
2. Circles meet when
3. r1 - r2
4. is at most d, and d is at most r1 + r2.

Answer: (c) 2

Explanation:
1. The circle on the diameter is \( x(x-1)+y(y-8)=0\Rightarrow x^2+y^2-x-8y=0 \).
2. Subtracting from C gives the common chord \( -5x-11=0\Rightarrow x=-\frac{11}5 \).
3. Putting it into the new circle: \( y^2-8y+7.04=0 \) with discriminant \( 64-28.16>0 \), so two real points.

Teacher's Note:
1. Subtract the two circles to get the radical axis.
2. Check the discriminant of the resulting quadratic.

Answer: (a) \( x^2+y^2-8x-8y+24=0 \)

Explanation:
1. Centre \( (h,h) \). Touching \( x+y=4 \): \( \frac{|2h-4|}{\sqrt2}=2\sqrt2\Rightarrow|2h-4|=4\Rightarrow h=4 \) or \( 0 \).
2. \( x+y>4 \) selects \( h=4 \).
3. \( (x-4)^2+(y-4)^2=8\Rightarrow x^2+y^2-8x-8y+24=0 \).

Teacher's Note:
1. Distance from the centre to the tangent line equals the radius.
2. Use the inequality to reject one of the two centres.

Answer: (b) \( (4\sqrt3+3\sqrt7)x-3y=18+6\sqrt{21} \)

Explanation:
1. A chord of length 8 is at distance \( \sqrt{25-16}=3 \) from the origin. Line: \( y-2=m(x-2\sqrt3) \), with \( \frac{|2-2\sqrt3m|}{\sqrt{1+m^2}}=3 \).
2. \( 3m^2-8\sqrt3m-5=0\Rightarrow m=\frac{4\sqrt3\pm3\sqrt7}3 \). The acute angle needs \( m>0 \): \( m=\frac{4\sqrt3+3\sqrt7}3 \).
3. The line: \( (4\sqrt3+3\sqrt7)x-3y=18+6\sqrt{21} \).

Teacher's Note:
1. Chord length gives its distance from the centre.
2. Positive slope means an acute angle with the x-axis.

Answer: (b) \( a\in[2,3],\ b\in[-3,2] \)

Explanation:
1. A point on the segment AB has coordinates between those of the end points.
2. \( a \) lies between 2 and 3 and \( b \) between \( -3 \) and 2.

Teacher's Note:
1. The chord AB is the straight segment between the end points.
2. Each coordinate varies between the end point values.

Answer: (a) \( (-\infty,-8)\cup(-2,6)\cup(6,\infty) \)

Explanation:
1. Outside the first: \( a^2+16+10a>0\Rightarrow(a+8)(a+2)>0\Rightarrow a\lt -8 \) or \( a>-2 \).
2. Outside the second: \( a^2-12a+36>0\Rightarrow(a-6)^2>0\Rightarrow a\ne6 \).
3. Combining: \( (-\infty,-8)\cup(-2,6)\cup(6,\infty) \).

Teacher's Note:
1. A point is outside when S1 is positive.
2. Intersect the two conditions.

Answer: (a) (25, 32)

Explanation:
1. Interior: \( 4+16-12-40+\lambda\lt 0\Rightarrow\lambda\lt 32 \).
2. Centre \( (3,5) \), \( r^2=34-\lambda \). Not cutting the axes: \( r\lt 3\Rightarrow34-\lambda\lt 9\Rightarrow\lambda>25 \).
3. So \( 25\lt \lambda\lt 32 \).

Teacher's Note:
1. Interior means S1 is negative.
2. Not meeting the axes means r is less than the distance from the centre to each axis.

Answer: (d) \( \left(\frac{2\pi}3,\frac{4\pi}3\right) \)

Explanation:
1. Interior: \( (1+\cos\theta)^2+\sin^2\theta\lt 1\Rightarrow2+2\cos\theta\lt 1\Rightarrow\cos\theta\lt -\frac12 \).
2. In \( [0,2\pi] \) this holds for \( \theta\in\left(\frac{2\pi}3,\frac{4\pi}3\right) \).

Teacher's Note:
1. Substitute the point and use sin squared + cos squared = 1.
2. Solve the trigonometric inequality on the circle.