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

2. A point $P$ moves in such a way that the ratio of its distance form two coplanar points is always a fixed number $(\ne 1)$ then its locus is

(a) Straight line

(b) circle

(c) Parabola

(d) a pair of Straight lines

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

Show Answer

Solution :

Given equation of circle in $x^2+y^2-4x-2y-20=0$

${\mathrm{S}}_1={10}^2+7^2-40-14-20>0$

$\therefore \mathrm{P}(10,7)$ lies outside the circle.

$\mathrm{PB}=\mathrm{PC}+\mathrm{CB}$

$\begin{array}{rl}& =r+\sqrt{8^2+6^2}\\ & =r+10\\ & =\sqrt{4+1+20}+10\\ & =5+10\\ & =15\end{array}$

4. If one end of a diameter of the circle $x^2+y^2-4x-6y+11=0$ be $(3,4)$, then the other end is

(a)$(0,0)$

(b)$(1,1)$

(c)$(1,2)$

(d) $(2,1)$.

Show Answer

Solution :

Centre of the circle is $(2,3)$ one end of the diameter is $(3,4)$. Since centre is the mid point of diameter

$\therefore \frac{\alpha +3}{2}=2\mathrm{\&}\frac{\beta +4}{2}=3$

$\alpha =1,\beta =2$

$\therefore$ Other end points is $(1,2)$.

5. $f(x,y)=x^2+y^2+2ax+2by+c=0$ 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

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 $x^2+y^2-10x-14y-51=0$ then G.M of $p$ and $q$ is equal to

(a) $2\sqrt{11}$

(b) $5\sqrt{5}$

(c) 13

(d) None of these

Show Answer

Solution :

The centre $\mathrm{C}$ of the circle is $(5,7)$ and the radius is $\sqrt{25+49+51}=\sqrt{125}=5\sqrt{5}$

$\begin{array}{rl}& \mathrm{PC}=\sqrt{{12}^2+5^2}=\sqrt{169}=13\\ & \begin{array}{r}\therefore \mathrm{p}=\mathrm{PB}\mathrm{PC}+\mathrm{CB}\mathrm{q}=\mathrm{PA}=\mathrm{PC}-\mathrm{CA}\\ =13+5\sqrt{5}=13-5\sqrt{5}\end{array}\\ & \begin{array}{rl}& \mathrm{GM}\text{ of }\mathrm{p}\mathrm{\&}\mathrm{q}=\sqrt{\mathrm{pq}}=\sqrt{(13+5\sqrt{5})(13-5\sqrt{5})}\\ & =\sqrt{{13}^2-(5\sqrt{5}{)}^2}\\ & =\sqrt{169-125}\end{array}\end{array}$

$\begin{array}{rl}& =\sqrt{44}\\ & =2\sqrt{11}\end{array}$

Examples

7. If $(1+\alpha x{)}^n=1+8x+24x^2+\dots \dots \dots ..$ 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 .

Show Answer

Solution :

$(1+\alpha x{)}^n=1+8x+24x^2+\dots \dots \dots \dots \dots$

$1+n\alpha x+n_{C_2}(\alpha x{)}^2+\dots \dots \dots \dots ......1+8x+24x^2+\dots \dots \dots \dots \dots$

$n\alpha x=8x{n}_{C_2}(\alpha x{)}^2=24x^2$

$n\alpha =8\frac{n(n-1)}{2}{\alpha }^2=24$ $\frac{n\alpha (n\alpha -\alpha )}{2}=24$

$\frac{(8-\alpha )}{2}=3$

$\frac{8-\alpha =6}{\alpha =2}\therefore n=4$

$\mathrm{P}(\alpha ,\mathrm{n})=\mathrm{P}(2,4)$ 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 $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x})f(\mathrm{y})\mathrm{\forall }\mathrm{x},\mathrm{y},f(1)=2$ and ${\alpha }_{\mathrm{n}}=\mathrm{f}(\mathrm{n}),\mathrm{n}\in \mathrm{N}$ then the equation of the circle having $({\alpha }_1,{\alpha }_2)$ and $({\alpha }_3,{\alpha }_4)$ 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) $(\mathrm{x}-6)(\mathrm{x}-8)+(\mathrm{y}-5)(\mathrm{y}-6)=0$

9. If $A$ and $B$ are two points on the circle $x^2+y^2-4x+6y-3=0$ 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) $\beta \equiv (2-2\sqrt{2},-3+2\sqrt{2})$

Show Answer

Solution :

$x^2+y^2-4x+6y-3=0$

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 $\frac{PA}{PB}=k$, then $k$ will not be equal to

(a) 0

(b) 1

(c) 2

(d) None of these

2. If the line $hx+ky=1$ 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) $\frac{1}{\sqrt{\mathrm{a}}}$

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) $x^2+y^2-6x+\frac{2y}{\sqrt{3}}+9=0$

(b) $x^2+y^2-6x-\frac{2y}{\sqrt{3}}+9=0$

(c) $x^2+y^2+6x+\frac{2y}{\sqrt{3}}+9=0$

(d) None of these

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) $9({\mathrm{x}}^2+{\mathrm{y}}^2)=4{\mathrm{a}}^2$

(b) $9(x^2+y^2)=a^2$

(c) $9({\mathrm{x}}^2+{\mathrm{y}}^2)=2{\mathrm{a}}^2$

(d) $9({\mathrm{x}}^2+{\mathrm{y}}^2)=8{\mathrm{a}}^2$

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) $x^2+2by=b^2+a^2$

(b) $x^2-2by=b^2+a^2$

(c) $x^2+2by=a^2-b^2$

(d) $x^2-2by=b^2-a^2$

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

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

8. The number of the points on the circle $x^2+y^2-4x-10y+13=0$ which are at a distance 1 from the point $(-3,2)$ is

(a) 1

(b) 2

(c) 3

(d) None of these

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) $x+y=1$

10. The area of the triangle formed by joining the origin to the points of intersection of the line $x\sqrt{5}+2y$ $=3\sqrt{5}$ and circle $x^2+y^2=10$ is

(a) 3

(b) 4

(c) 5

(d) 6

11. If $(-3,2)$ lies on the circle $x^2+y^2+2gx+2fy+c=0$ which is concentric with the circle $x^2+y^2+6x+8y-$ $5=0$ then $\mathrm{c}$ is

(a) $11$

(b) $-11$

(c) $24$

(d) None of these

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) $(x-p{)}^2=4qy$

(b) $(x-q{)}^2=4qy$

(c) $(\mathrm{y}-\mathrm{p}{)}^2=4\mathrm{qx}$

(d) $(\mathrm{y}-\mathrm{q}{)}^2=4\mathrm{px}$

13. If the lines $2x+3y+1=0$ and $3x-y-4=0$ lie along diameters of a circle of circmference $10\pi$, then the equation of the circle is

(a) $x^2+y^2-2x+2y-23=0$

(b) $x^2+y^2+2x+2y-23=0$

(c) $x^2+y^2-2x-2y-23=0$

(d) $x^2+y^2+2x-2y-23=0$

14. The lines $2x-3y=5$ and $3x-4y=7$ are diameters of a circle having area as 154 sq.unit, then the equation of the circle is

(a) $x^2+y^2+2x-2y=62$

(b) $x^2+y^2+2x-2y=47$

(c) $x^2+y^2-2x+2y=47$

(d) $x^2+y^2-2x+2y=62$

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) $(x-5/2{)}^2+(y+3{)}^2=\frac{61}{4}$