CIRCLE-3 (Problem Solving)
1. The number of points $(\mathrm{x}, \mathrm{y})$ having integral coordinates satisfying the condition $\mathrm{x}^{2}+\mathrm{y}^{2}<25$ is
(a) 90
(b) 81
(c) 80
(d) 69
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Solution :
Since $x^{2}+y^{2}<25$ and $x$ and $y$ are integers, the possible values of $x \& y \in(0, \pm 1, \pm 2, \pm 3, \pm 4)$, thus $x$ and $y$ can be chosen in 9 ways each and $\left(x _{1} y\right)$ can be $9 \times 9=81$ ways. But $( \pm 3, \pm 4)$ $( \pm 4, \pm 3)( \pm 4, \pm 4)$ does not satisfy so we must exclude these points $3 \times 4=12$ ways.
Hence the number of permissible values are $81-12=69$.
2. A point $P$ moves in such a way that the ratio of its distance form two coplanar points is always a fixed number $(\neq 1)$ then its locus is
(a) Straight line
(b) circle
(c) Parabola
(d) a pair of Straight lines
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Solution :
Let two coplanar points be $(0,0)$ and $(a, 0)$ according to the question we get
$\frac{\mathrm{AP}}{\mathrm{BP}}=\lambda($ constant)
$\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{(x-a)^{2}+y^{2}}}=\lambda$
$\mathrm{x}^{2}+\mathrm{y}^{2}=\lambda^{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ax}+\mathrm{a}^{2}\right)$
$\mathrm{x}^{2}+\mathrm{y}^{2}+\frac{\lambda^{2}}{1-\lambda^{2}}\left(2 \mathrm{ax}-\mathrm{a}^{2}\right)=0$
represents equation of a circle
3. The greatest distance of the point $P(10,7)$ from the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-2 \mathrm{y}-20=0$ is
(a) 10
(b) 15
(c) 5
(d) None of these
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Solution :
Given equation of circle in $x^{2}+y^{2}-4 x-2 y-20=0$
$\mathrm{S} _{1}=10^{2}+7^{2}-40-14-20>0$
$\therefore \mathrm{P}(10,7)$ lies outside the circle.
$\mathrm{PB}=\mathrm{PC}+\mathrm{CB}$
$ \begin{aligned} & =r+\sqrt{8^{2}+6^{2}} \\ & =r+10 \\ & =\sqrt{4+1+20}+10 \\ & =5+10 \\ & =15 \end{aligned} $
4. If one end of a diameter of the circle $x^{2}+y^{2}-4 x-6 y+11=0$ be $(3,4)$, then the other end is
(a)$(0,0)$
(b)$(1,1)$
(c)$(1,2)$
(d) $(2,1)$.
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Solution :
Centre of the circle is $(2,3)$ one end of the diameter is $(3,4)$. Since centre is the mid point of diameter
$\therefore \frac{\alpha+3}{2}=2 \& \frac{\beta+4}{2}=3$
$\alpha=1, \beta=2$
$\therefore$ Other end points is $(1,2)$.
5. $f(x, y)=x^{2}+y^{2}+2 a x+2 b y+c=0$ represents a circle. If $f(x, 0)=0$ has equal roots each being 2 and $f(0, y)=0$ has 2 and 3 as its roots then the centre of circle is
(a) $(2,5 / 2)$
(b) date are not sufficient
(c) $(-2,-5 / 2)$
(d) data are inconsistent
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Solution :
$ \begin{aligned} f(x, 0)=0 \Rightarrow \quad x^{2}+2 a x+c=0 & =(x-2)^{2} \\ \therefore c=4 . & \& 2 a=-4 . \Rightarrow a=-2 \\ f(0, y)=0 \Rightarrow \quad y^{2}+2 b y+c=0 & =(y-2)(y-3) \\ & =y^{2}-5 y+6 \\ \therefore b=-5 / 2 \& c & =6 \end{aligned} $
$\therefore \mathrm{c}$ is not unique so data are inconsistent
6. If $\mathrm{p}$ and $\mathrm{q}$ are the largest distance and the shortest distance respectively of the point $(-7,2)$ from any point $(\alpha, \beta)$ on the curve whose equation is $x^{2}+y^{2}-10 x-14 y-51=0$ then G.M of $p$ and $q$ is equal to
(a) $2 \sqrt{11}$
(b) $5 \sqrt{5}$
(c) 13
(d) None of these
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Solution :
The centre $\mathrm{C}$ of the circle is $(5,7)$ and the radius is $\sqrt{25+49+51}=\sqrt{125}=5 \sqrt{5}$
$ \begin{aligned} & \mathrm{PC}=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \\ & \begin{aligned} \therefore \mathrm{p}=\mathrm{PB} \quad \mathrm{PC}+\mathrm{CB} \quad \mathrm{q}=\mathrm{PA}=\mathrm{PC}-\mathrm{CA} \\ =13+5 \sqrt{5} \quad=13-5 \sqrt{5} \end{aligned} \\ & \begin{aligned} & \mathrm{GM} \text { of } \mathrm{p} \& \mathrm{q}=\sqrt{\mathrm{pq}} \quad=\sqrt{(13+5 \sqrt{5})(13-5 \sqrt{5})} \\ &=\sqrt{13^{2}-(5 \sqrt{5})^{2}} \\ &=\sqrt{169-125} \end{aligned} \end{aligned} $
$ \begin{aligned} & =\sqrt{44} \\ & =2 \sqrt{11} \end{aligned} $
Examples
7. If $(1+\alpha x)^{n}=1+8 x+24 x^{2}+………..$ and a line through $\mathrm{P}(\alpha, n)$ cuts the circle $x^{2}+y^{2}=4$ in A and B, then PA.PB is
(a) 4
(b) 8
(c) 16
(d) 32 .
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Solution :
$(1+\alpha x)^{n}=1+8 x+24 x^{2}+\ldots \ldots \ldots \ldots \ldots $
$1+n \alpha x+n_{C_{2}}(\alpha x)^{2}+\ldots \ldots \ldots \ldots . . . . . .1+8 x+24 x^{2}+\ldots \ldots \ldots \ldots \ldots $
$n \alpha x=8 x n_{C_{2}}(\alpha x)^{2}=24 x^{2} $
$n \alpha=8 \frac{n(n-1)}{2} \alpha^{2}=24 $ $ \frac{n \alpha(n \alpha-\alpha)}{2}=24 $
$ \frac{(8-\alpha)}{2}=3$
$ \frac{8-\alpha=6}{\alpha=2} \therefore n=4$
$\mathrm{P}(\alpha, \mathrm{n})=\mathrm{P}(2,4)$ and $\mathrm{PT}$ is a tangent of length 4
We know that
$\mathrm{PT}^{2}=\mathrm{PA} . \mathrm{PB}=4^{2}=16$
(Secant tangent theorem)
8. If $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y}) \forall \mathrm{x}, \mathrm{y}, f(1)=2$ and $\alpha _{\mathrm{n}}=\mathrm{f}(\mathrm{n}), \mathrm{n} \in \mathrm{N}$ then the equation of the circle having $\left(\alpha _{1}, \alpha _{2}\right)$ and $\left(\alpha _{3}, \alpha _{4}\right)$ as the ends of its one diameter is
(a) $(x-2)(x-8)+(y-4)(y-16)=0$
(b) $(x-4)(x-8)+(y-2)(y-16)=0$
(c) $(\mathrm{x}-2)(\mathrm{x}-16)+(\mathrm{y}-4)(\mathrm{y}-8)=0$
(d) $(\mathrm{x}-6)(\mathrm{x}-8)+(\mathrm{y}-5)(\mathrm{y}-6)=0$
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Solution :
$ \begin{aligned} & f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y}) \\ & f(2)=f(1+1)=f(1) f(1)=2^{2}=\alpha _{2} \\ & \alpha _{3}=f(3)=2^{3}, f(4)=2^{4}=\alpha _{4} \\ & \left(\alpha _{1}, \alpha _{2}\right)=(2,4) \&\left(\alpha _{3}, \alpha _{4}\right)=(8,16) \end{aligned} $
Equation of circle in diameter form
$ \begin{aligned} & \left(\mathrm{x}-\mathrm{x} _{1}\right)\left(\mathrm{x}-\mathrm{x} _{2}\right)+\left(\mathrm{y}-\mathrm{y} _{1}\right)\left(\mathrm{y}-\mathrm{y} _{2}\right)=0 \\ & (\mathrm{x}-2)(\mathrm{x}-8)+(\mathrm{y}-4)(\mathrm{y}-16)=0 \end{aligned} $
9. If $A$ and $B$ are two points on the circle $x^{2}+y^{2}-4 x+6 y-3=0$ which are farthest and nearest respectively from the point $(7,2)$ then
(a) $ \mathrm{A} \equiv(2-2 \sqrt{2},-3-2 \sqrt{2})$
(b) $ \mathrm{A} \equiv(2+2 \sqrt{2},-3+2 \sqrt{2})$
(c) $ \beta \equiv(2+2 \sqrt{2},-3+2 \sqrt{2})$
(d) $ \beta \equiv(2-2 \sqrt{2},-3+2 \sqrt{2})$
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Solution :
$x^{2}+y^{2}-4 x+6 y-3=0$
Centre of the circle is $(2,-3)$
Radius of the circle is $=\sqrt{2^{2}+3^{3}+3}=\sqrt{16}=4$
$S _{1}=49+4-28+12-3$
$=65-31$
$=34>0$
$\therefore$ point $(7,2)$ lies outside the circle
$\mathrm{PC}=\sqrt{(7-2)^{2}+(2+3)^{2}}$
$=\sqrt{5^{2}+5^{2}}$
$=\sqrt{50}=5 \sqrt{2}$
$\mathrm{CA}=\mathrm{CB}=\mathrm{r}=4$
Farthest point
$\mathrm{PA}=\mathrm{PC}+\mathrm{CA}$
$ =5 \sqrt{2}+4 $
Nearest point
$\mathrm{PB}=\mathrm{PC}-\mathrm{CB}$
$ =5 \sqrt{2}-4 $
By tinding Point of pnteraction
$\therefore$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are
$\mathbf{A}((2-2 \sqrt{2}),(-3-2 \sqrt{2}))$ and $\mathrm{B}(2+2 \sqrt{2},-3+2 \sqrt{2})$
Practice questions
1. The points $A$ and $B$ in a plane are such that for all points $P$ lies on circle Satisfying $\frac{P A}{P B}=k$, then $k$ will not be equal to
(a) 0
(b) 1
(c) 2
(d) None of these
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Answer: (b)2. If the line $h x+k y=1$ touches $x^{2}+y^{2}=a^{2}$ then the locus of the point $(h, k)$ is a circle of radius
(a) a
(b) $1 / \mathrm{a}$
(c) $\sqrt{\mathrm{a}}$
(d) $\frac{1}{\sqrt{\mathrm{a}}}$
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Answer: (b)3. Equation of incircle of equilateral triangle $\mathrm{ABC}$ where $\mathrm{B}(2,0), \mathrm{C}(4,0)$ and $\mathrm{A}$ lies in the fourth quadrant is
(a) $x^{2}+y^{2}-6 x+\frac{2 y}{\sqrt{3}}+9=0$
(b) $x^{2}+y^{2}-6 x-\frac{2 y}{\sqrt{3}}+9=0$
(c) $x^{2}+y^{2}+6 x+\frac{2 y}{\sqrt{3}}+9=0$
(d) None of these
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Answer: (a)4. A variable circle having chord of radius ’ $\mathrm{a}$ ’ passes through origin meets the coordinate axes in points $\mathrm{A}$ and $\mathrm{B}$. locus of centroid of triangle $\mathrm{OAB}$, ’ $\mathrm{O}$ ’ being the origin is
(a) $9\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=4 \mathrm{a}^{2}$
(b) $9\left(x^{2}+y^{2}\right)=a^{2}$
(c) $\quad 9\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=2 \mathrm{a}^{2}$
(d) $9\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=8 \mathrm{a}^{2}$
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Answer: (a)5. The locus of the centre of the circle which cuts a chord of length 2a from the positive $x$-axis and passes through a point on positive $y$-axis distant $b$ from the origin is
(a) $x^{2}+2 b y=b^{2}+a^{2}$
(b) $ x^{2}-2 b y=b^{2}+a^{2}$
(c) $x^{2}+2 b y=a^{2}-b^{2}$
(d) $ x^{2}-2 b y=b^{2}-a^{2}$
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Answer: (c)6. The number of circle having radius 5 and passing through the points $(-2,0)$ and $(4,0)$ is
(a) one
(b) two
(c) four
(d) infinite
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Answer: (b)7. The equation of the smallest circle passing through the intersection of the line $x+y=1$ and the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=9$ is
(a) $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}+\mathrm{y}-8=0$
(b) $\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-\mathrm{y}-8=0$
(c) $\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-\mathrm{y}+8=0$
(d) None of these
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Answer: (b)8. The number of the points on the circle $x^{2}+y^{2}-4 x-10 y+13=0$ which are at a distance 1 from the point $(-3,2)$ is
(a) 1
(b) 2
(c) 3
(d) None of these
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Answer: (d)9. The locus of the mid-point of a chord of the circle $x^{2}+y^{2}=4$ which subtends a right angle at the origin is
(a) $x+y=2$
(b) $x^{2}+y^{2}=1$
(c) $x^{2}+y^{2}=2$
(d) $x+y=1$
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Answer: (c)10. The area of the triangle formed by joining the origin to the points of intersection of the line $x \sqrt{5}+2 y$ $=3 \sqrt{5}$ and circle $x^{2}+y^{2}=10$ is
(a) 3
(b) 4
(c) 5
(d) 6
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Answer: (c)11. If $(-3,2)$ lies on the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ which is concentric with the circle $x^{2}+y^{2}+6 x+8 y-$ $5=0$ then $\mathrm{c}$ is
(a) $11$
(b) $-11$
(c) $24$
(d) None of these
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Answer: (b)12. A variable circle passes through the fixed point $A(p, q)$ and touches $x$-axis. The locus of the other end of the diameter through A is
(a) $(x-p)^{2}=4 q y$
(b) $(x-q)^{2}=4 q y$
(c) $(\mathrm{y}-\mathrm{p})^{2}=4 \mathrm{qx}$
(d) $(\mathrm{y}-\mathrm{q})^{2}=4 \mathrm{px}$
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Answer: (a)13. If the lines $2 x+3 y+1=0$ and $3 x-y-4=0$ lie along diameters of a circle of circmference $10 \pi$, then the equation of the circle is
(a) $x^{2}+y^{2}-2 x+2 y-23=0$
(b) $x^{2}+y^{2}+2 x+2 y-23=0$
(c) $x^{2}+y^{2}-2 x-2 y-23=0$
(d) $x^{2}+y^{2}+2 x-2 y-23=0$
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Answer: (a)14. The lines $2 x-3 y=5$ and $3 x-4 y=7$ are diameters of a circle having area as 154 sq.unit, then the equation of the circle is
(a) $x^{2}+y^{2}+2 x-2 y=62$
(b) $x^{2}+y^{2}+2 x-2 y=47$
(c) $x^{2}+y^{2}-2 x+2 y=47$
(d) $ x^{2}+y^{2}-2 x+2 y=62$
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Answer: (c)15. The equation of circle which passes through the origin and cuts off intercepts 5 and 6 from the positive parts of the axes respectively, is
(a) $(x+5 / 2)^{2}+(y+3)^{2}=\frac{61}{4}$
(b) $(x-5 / 2)^{2}+(y-3)^{2}=\frac{61}{4}$
(c) $(x-5 / 2)^{2}-(y-3)^{2}=\frac{61}{4}$
(d) $(x-5 / 2)^{2}+(y+3)^{2}=\frac{61}{4}$
