Example 1

Find the equation of the image of the circle $x^2+y^2+16x-24y+183=0$ by the line mirror $4x+7y+13=0$

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Solution:

The given circle and line are

${\mathrm{x}}^2+{\mathrm{y}}^2+16\mathrm{x}-24\mathrm{y}+183=0$______________(1) and $4x+7y+13=0$ _________________(2)

Centre and radius if the circle are

$(-8,12)$ and $\sqrt{64+144-183}=\sqrt{25}=5$ respectively.

Equation of line ${\mathrm{C}}_1{\mathrm{C}}_2$ is $7\mathrm{x}-4\mathrm{y}+\mathrm{k}=0$ it passes through $(-8,12)$

$\therefore -56-48+\mathrm{k}=0$

$\mathrm{k}=104$

Equation of line ${\mathrm{C}}_1{\mathrm{C}}_2$ is $7\mathrm{x}-4\mathrm{y}+104=0$

To get the coordinates of M. Solve the equation

(2) & (3)

(2) $\times 4+(3)\times 7$

$16\mathrm{x}+28\mathrm{y}+52=0$

$\begin{array}{c}49x-28y+728=0\\ 65x+780=0\end{array}$

$\mathrm{x}=-12$

put the value of $x$ in (2) we get

$-48+7y+13=0$

$7\mathrm{y}=35$

$\mathrm{y}=5$

$\therefore$ coordinate of $\mathrm{M}$ is $(-12,5)$

$M$ is the midpoint of ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$

$\therefore -12={\displaystyle \frac{\mathrm{h}-8}{2}}\Rightarrow \mathrm{h}=-16$

$5={\displaystyle \frac{\mathrm{k}+12}{2}}\Rightarrow \mathrm{K}=-2$

$\therefore$ Equation of imaged circle is

$(x+16{)}^2+(y+2{)}^2=25$

$x^2+y^2+32x+4y+235=0$

1. The circle passing through the point $(-1,0)$ and touching the $y$-axis at $(0,2)$ also passes through the point

(a) $\left({\displaystyle \frac{-3}{2}}10\right)$

(b) $({\displaystyle \frac{-5}{2}},12)$

(c) $({\displaystyle \frac{-3}{2}},{\displaystyle \frac{5}{2}})$

(d) $(-4,0)$

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Solution:

Family of circles passing through a point $(0,2)$ and touching line $\mathrm{x}=0$ (y-axis) is

$(x-0{)}^2+(y-2{)}^2+\lambda x=0$

It passes through $(-1,0)$

$\therefore 1+4-\lambda =0$

$\therefore \lambda =5$

$\therefore$ equation of circle is

${\mathrm{x}}^2+{\mathrm{y}}^2+5\mathrm{x}-4\mathrm{y}+4=0$

It also passes through $\mathrm{A}(-{\mathrm{x}}_1,0)$

$\therefore {\mathrm{x}}_1^2-5{\mathrm{x}}_1+4=0$

$({\mathrm{x}}_1-4)({\mathrm{x}}_1-1)=0$

${\mathrm{x}}_1=4,{\mathrm{x}}_1=1$

$\therefore$ it also passes through $\mathrm{A}(-4,0)$

2. Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the centre, angles of ${\displaystyle \frac{\pi }{\mathrm{k}}}$ and ${\displaystyle \frac{2\pi }{\mathrm{k}}}$, where $\mathrm{k}>0$ then the value of $[\mathrm{k}]$ is.

( $[\mathrm{k}]$ denotes the greatest integer)

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Solution:

Let ${\displaystyle \frac{\pi }{2\mathrm{k}}}=\alpha ={\displaystyle \frac{1}{2}}\mathrm{\angle }\mathrm{AOB}=\mathrm{\angle }\mathrm{AOM}$

Then $\mathrm{\angle }\mathrm{CON}=2\alpha$

${\mathrm{In}}_{\mathrm{\Delta }}\mathrm{AOM}$

$\cos \alpha ={\displaystyle \frac{\mathrm{X}}{2}}$

In $\mathrm{\Delta }\mathrm{CON}$

$\cos 2\alpha ={\displaystyle \frac{\sqrt{3}+1-\mathrm{x}}{2}}$

$\cos 2\alpha =2{\cos }^2\alpha -1$

$\cos 2\alpha =2{\displaystyle \frac{x^2}{4}}-1$

$\therefore {\displaystyle \frac{\sqrt{3}+1-\mathrm{x}}{2}}={\displaystyle \frac{{\mathrm{x}}^2}{2}}-1$

$\begin{array}{rl}& \sqrt{3}+1-x=x^2-2\\ & x^2+x-\sqrt{3}-3=0\\ & \therefore x={\displaystyle \frac{-1\pm \sqrt{1+4(3+\sqrt{3})}}{2}}\\ & ={\displaystyle \frac{-1\pm \sqrt{13+4\sqrt{3}}}{2}}\\ & ={\displaystyle \frac{-1\pm (2\sqrt{3}+1)}{2}}(\therefore 13+4\sqrt{3}=(2\sqrt{3}+1{)}^2)\\ & x={\displaystyle \frac{-1+2\sqrt{3}+1}{2}}\\ & x=\sqrt{3}\\ & \cos \alpha ={\displaystyle \frac{\sqrt{3}}{2}}=\cos {\displaystyle \frac{\pi }{6}}\\ & \alpha ={\displaystyle \frac{\pi }{6}}\\ & \text{ Required angle }={\displaystyle \frac{\pi }{k}}=2\alpha ={\displaystyle \frac{\pi }{3}}\end{array}$

3. Let $\mathrm{ABC}$ and ${\mathrm{AB}}^{\prime }$ be two non-congruent triangles with sides $\mathrm{AB}=4,\mathrm{AC}={\mathrm{AC}}^{\prime }=2\sqrt{2}$ and angle $\beta ={30}^{\circ }$. The absolute value of the difference between the areas of these triangles is

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Solution:

Draw circle through $\mathrm{AC}{\mathrm{C}}^{\prime }$ and $\mathrm{AB}$ intersect the circle at $\mathrm{P}$

$\begin{array}{rl}& {\mathrm{In}}_{\mathrm{\Delta }}\mathrm{ABD}\\ & ={\displaystyle \frac{\mathrm{AD}}{\mathrm{AB}}}=\sin 30\\ & {\displaystyle \frac{\mathrm{AD}}{4}}={\displaystyle \frac{1}{2}}\\ & \therefore \mathrm{AD}=2=\mathrm{DC}={\mathrm{C}}^{\prime }\mathrm{D}\end{array}$

Difference of areas of $\mathrm{△}\mathrm{ABC}$ and $\mathrm{△}{\mathrm{ABC}}^{\prime }$ is $\mathrm{\Delta }{\mathrm{ACC}}^{\prime }$

$\therefore \mathrm{ar}\left(\mathrm{\Delta }{\mathrm{ACC}}^{\prime }\right)={\displaystyle \frac{1}{2}}\times 4\times 2=4$ sq.u.

4. The centres of two circles ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ each of unit radius are at a distance of 6 units from each other. Let $\mathrm{P}$ be the mid-point of the line segment joining the centres of ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ and $\mathrm{C}$ be a circle touching circles ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ externally. If a common tangents to ${\mathrm{C}}_1$ and $\mathrm{C}$ passing through $\mathrm{P}$ is also a common targent to ${\mathrm{C}}_2$ and ${\mathrm{C}}_1$ then the radius of the circle $\mathrm{C}$ is

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Solution:

In $\mathrm{\Delta }{\mathrm{CPC}}_2$

${\mathrm{CP}}^2={\left({\mathrm{CC}}_2\right)}^2-{\left({\mathrm{C}}_2\mathrm{P}\right)}^2$

${\mathrm{k}}^2=(\mathrm{r}+1{)}^2-9$

${\mathrm{k}}^2={\mathrm{r}}^2+2\mathrm{r}-8$____________________(1)

In $\mathrm{△}{\mathrm{PQC}}_2$

$\mathrm{PQ}{}^2=3^2-1^2$

$=8$

$\therefore$ In ${}_{\mathrm{\Delta }}\mathrm{CPQ}$

${\mathrm{k}}^2={\mathrm{r}}^2+8$____________________(2)

From (1) & (2)

${\mathrm{F}}^2+8={\mathrm{r}}^2+2\mathrm{r}-8$

$2\mathrm{r}=16$

$\therefore \mathrm{r}=8$

5. A straight line through the vertex $\mathrm{P}$ of a triangle $\mathrm{PQR}$ intersects the side $\mathrm{QR}$ at the point $\mathrm{S}$ and the circumcircle of the triangle $\mathrm{PQR}$ at the point $\mathrm{T}$. If $\mathrm{S}$ is not the centre of the circumcircle then

(a) ${\displaystyle \frac{1}{\mathrm{PS}}}+{\displaystyle \frac{1}{\mathrm{ST}}}<{\displaystyle \frac{2}{\sqrt{\mathrm{QS}.\mathrm{SR}}}}$

(b) ${\displaystyle \frac{1}{\mathrm{PS}}}+{\displaystyle \frac{1}{\mathrm{ST}}}>{\displaystyle \frac{2}{\sqrt{\mathrm{QS}.\mathrm{SR}}}}$

(c) ${\displaystyle \frac{1}{\mathrm{PS}}}+{\displaystyle \frac{1}{\mathrm{ST}}}<{\displaystyle \frac{4}{\mathrm{QR}}}$

(d) ${\displaystyle \frac{1}{\mathrm{PS}}}+{\displaystyle \frac{1}{\mathrm{ST}}}>{\displaystyle \frac{4}{\mathrm{QR}}}$

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Solution:

Points $\mathrm{P},\mathrm{Q},\mathrm{T},\mathrm{R}$ are concyclic

$\therefore$ PS.ST $=$ QS.SR

$\Rightarrow {\displaystyle \frac{\mathrm{PS}+\mathrm{ST}}{2}}\ge \sqrt{\mathrm{PS}.\mathrm{ST}}(\mathrm{AM}\ge \mathrm{GM})$

$\therefore \mathrm{PT}\ge 2\sqrt{\mathrm{PS}.\mathrm{ST}}$

and ${\displaystyle \frac{1}{\mathrm{PS}}}+{\displaystyle \frac{1}{\mathrm{ST}}}\ge {\displaystyle \frac{2}{\sqrt{\mathrm{PSST}}}}={\displaystyle \frac{2}{\sqrt{\mathrm{QSSR}}}}$

Also, ${\displaystyle \frac{\mathrm{SQ}+\mathrm{SR}}{2}}\ge \sqrt{\mathrm{SQ}.\mathrm{SR}}$

$\Rightarrow {\displaystyle \frac{\mathrm{QR}}{2}}\ge \sqrt{\mathrm{SQ}.\mathrm{SR}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{\mathrm{SQ}\cdot SR}}}\ge {\displaystyle \frac{2}{\mathrm{QR}}}$

$\Rightarrow {\displaystyle \frac{2}{\sqrt{\mathrm{SQ}\cdot \mathrm{SR}}}}\ge {\displaystyle \frac{4}{\mathrm{QR}}}$

${\displaystyle \frac{1}{\mathrm{PS}}}+{\displaystyle \frac{1}{\mathrm{ST}}}\ge {\displaystyle \frac{2}{\sqrt{\mathrm{QS}.\mathrm{SR}}}}\ge {\displaystyle \frac{4}{\mathrm{QR}}}$

Answer: (d)

6. Let $\mathrm{ABCD}$ be a quadrilateral with area 18 , with side $\mathrm{AB}$ parallel to the side $\mathrm{CD}$ and $AB=2CD$. Let $AD$ be perpendicular to $AB$ and $CD$. If a circle is drawn inside the quadrilateral $\mathrm{ABCD}$ touching all the sides then its radius is

(a) 3

(b) 2

(c) $3/2$

(d) 1 .

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Solution:

$\mathrm{ABCD}$ is a trapezium $(\because$ AB parallel to $\mathrm{CD})$

$\therefore \mathrm{ar}(\mathrm{ABCD})={\displaystyle \frac{1}{2}}\times \mathrm{h}$ (sum of parallel sides)

$={\displaystyle \frac{1}{2}}\times 2\mathrm{r}(2\mathrm{a}+\mathrm{a})$

$18=\mathrm{r}\times 3\mathrm{a}$

ar $=6$

$\mathrm{CB}$ is a tangent to the circle

$\therefore$ equation of tangent is

$y={\displaystyle \frac{-2r}{a}}(x-2a)\Rightarrow 2rx+ay-4ar=0$

It is a tangent to the circle $(x-r{)}^2+(y-r{)}^2=r^2$

$\therefore \mathrm{r}=\left|{\displaystyle \frac{2{\mathrm{r}}^2+\mathrm{ar}-4\mathrm{ar}}{\sqrt{4{\mathrm{r}}^2+{\mathrm{a}}^2}}}\right|$

$r\sqrt{4r^2+a^2}=2r^2-3ar$

$\sqrt{4{\mathrm{r}}^2+{\mathrm{a}}^2}=2\mathrm{r}-3\mathrm{a}$

Squaring

$4r^2+a^2=4r^2+9a^2-12ar$

$12\mathrm{r}=8\mathrm{a}\Rightarrow 3\mathrm{r}=2\mathrm{a}$

ar $=6$

$\mathrm{r}={\displaystyle \frac{2\mathrm{a}}{3}}$

${\displaystyle \frac{2{\mathrm{a}}^2}{3}}=6$

${\mathrm{a}}^2=9$

$\mathrm{a}=\pm 3$

$\therefore \mathrm{r}=2$

Answer: (b)

7. The radius of the least circle passing through the point $(8,4)$ and cutting the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=40$ orthogonally is

(a) $\sqrt{5}$

(b) $\sqrt{7}$

(c) $2\sqrt{5}$

(d) $4\sqrt{5}$

8. $P$ is a point $(a,b)$ in the first quadrant. If the two circles which pass throngh $P$ and touch both the co-ordinate axes cut at right angles, then

(a) ${\mathrm{a}}^2-6\mathrm{ab}+{\mathrm{b}}^2=0$

(b) ${\mathrm{a}}^2+2\mathrm{ab}-{\mathrm{b}}^2=0$

(c) ${\mathrm{a}}^2-4\mathrm{ab}+{\mathrm{b}}^2=0$

(d) ${\mathrm{a}}^2-8\mathrm{ab}+{\mathrm{b}}^2=0$

9. A circle $S\equiv 0$ passes through the common points of family of circles $x^2+y^2+\lambda x-4y+3=0(\lambda \in R)$ and have minimum area then

(a) area of $S\equiv 0$ is $\pi$ sq.u

(b) radius of director circle of $S\equiv 0$ is $\sqrt{2}$

(c) Radius of director circle of $\mathrm{S}\equiv 0$ for $\mathrm{x}$-axis is 1 unit

(d) $\mathrm{S}\equiv 0$ never cuts $|2\mathrm{x}|=1$

10. Area of part of circle $x^2+y^2-4x-6y+12=0$ above the line $4x+7y-29=0$ is $\mathrm{\Delta }$, then $[\mathrm{\Delta }]=$ [.] is greatest integer function.

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Solution

Since line $4x+7y-29=0$

passes through the centre $(2,3)$ of the circle

$\therefore$ The line is a diameter of a circle with radius $\sqrt{4+9-12}=1$

$\therefore$ area of semi circle is $={\displaystyle \frac{1}{2}}\pi {\mathrm{r}}^2$

$\begin{array}{rl}& ={\displaystyle \frac{1}{2}}\pi \\ & ={\displaystyle \frac{3.14}{2}}=1.57\end{array}$

Hence $[\mathrm{\Delta }]=[1.57]=1$

EXERCISE:

1. The number of common tangents that can be drawn to the circles $x^2+y^2-4x-6y-3=0$ and

(a) 1

(b) 2

(c) 3

(d) 4

2. A variable chord is drawn through the origin to the circle $x^2+y^2-2ax=0$. The locus of the centre of the circle drawn on this chord as diameter is

(a) $x^2+y^2+ax=0$

(b) $x^2+y^2+ay=0$

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

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

3. If $O$ is the origin and $\mathrm{OP},\mathrm{OQ}$ are distinct tangents to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$, the circumcentre of the triangle OPQ is

(a) $(-\mathrm{g},-\mathrm{f})$

(b) $(\mathrm{g},\mathrm{f})$

(c) $(-\mathrm{f},-\mathrm{g})$

(d) none of these

4. Equation of the normal to the circle $x^2+y^2-4x+4y-17=0$ which passes through $(1,1)$ is

(a) $3x+2y-5=0$

(b) $3x+y-4=0$

(c) $3x+2y-2=0$

(d) $3\mathrm{x}-\mathrm{y}-8=0$

5. The equation of the circle touching the lines $|y|=x$ at a distance $\sqrt{2}$ unit from the origin is

(a) $x^2+y^2-4x+2=0$

(b) $x^2+y^2+4x-2=0$

(c) $x^2+y^2+4x+2=0$

(d) none of these

6. The shortest distance from the point $(2,-7)$ to the circle $x^2+y^2-14x-10y-151=0$ is

(a) 1

(b) 2

(c) 3

(d) 4

7. The equation of the image of the circle $(x-3)2+(y-2)2=1$ by the mirror $x+y=19$ is

(a) $(x-14{)}^2+(y-13{)}^2=1$

(b) $(x-15{)}^2+(y-14{)}^2=1$

(c) $(x-16{)}^2+(y-15{)}^2=1$

(d) $(x-17{)}^2+(y-16{)}^2=1$

8. If $P$ and $Q$ are two points on the circle $x^2+y^2-4x-4y-1$ which are farthest and nearest respectively from the point $(6,5)$, then

(a) $\mathrm{P}=(-{\displaystyle \frac{22}{5}},3)$

(b) $\mathrm{Q}=({\displaystyle \frac{22}{5}},{\displaystyle \frac{19}{5}})$

(c) $\mathrm{P}=({\displaystyle \frac{14}{3}},-{\displaystyle \frac{11}{5}})$

(d) $\mathrm{Q}=(-{\displaystyle \frac{14}{3}},-4)$

9. A circle of the coaxial system with limiting points $(0,0)$ and $(1,0)$ is

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

(b) $x^2+y^2-6x+3=0$

(c) $x^2+y^2=1$

(d) $x^2+y^2-2x+1=0$

10. If a variable circle touches externally two given circles, then the locus of the centre of the variable circle is

(a) a straight line

(b) a parabola

(c) an ellipse

(d) a hyperbola

PASSAGE

For each natural number $\mathrm{k}$, let ${\mathrm{C}}_{\mathrm{k}}$ denotes the circle with radius $\mathrm{k}$ units and centre at the origin. On the ${\mathrm{C}}_{\mathrm{k}}$, a particle moves $\mathrm{k}$ units in the counter clockwise direction. After completing its motion on ${\mathrm{C}}_{\mathrm{k}}$, the particle moves to ${\mathrm{C}}_{\mathrm{k}+\ell }$ in some well defined manner, where $\ell >0$. The motion of the particle continues in this manner.

On the basis of above information, answer the following questions:

11. Let $\ell =1$, the particle starts at $(1,0)$. If the particle crossing the positive direction of the $\mathrm{x}$-axis for the first time on the circle ${\mathrm{C}}_{\mathrm{n}}$, then $\mathrm{n}$ is equal to

(a) 3

(b) 5

(c) 7

(d) 8

12. If $\mathrm{k}\in \mathrm{N}$ and $\ell =1$, the particle starts $(-1,0)$ the particle cross $\mathrm{x}$-axis again at

(a) $(3,0)$

(b) $(1,0)$

(c) $(4,0)$

(d) $(2,0)$

13. If $\mathrm{k}\in \mathrm{N}$ and $\ell =1$, the particle moves in the radial direction from circle ${\mathrm{C}}_{\mathrm{k}}$ to ${\mathrm{C}}_{\mathrm{k}}+1$. If particle starts form the point $(-1,0)$, then

(a) it will cross the + ve $y$-axis at $(0,4)$

(b) it will cross the - ve $y$-axis at $(0,-4)$

(c) it will cross the $+vey$-axis at $(0,5)$

(d) it will cross the - ve y-axis at $(0,-5)$

14. If $\mathrm{k}\in \mathrm{N}$ and $\ell =\mathrm{R}$, particle moves tangentially from the circle ${\mathrm{C}}_{\mathrm{k}}$ to ${\mathrm{C}}_{\mathrm{k}+1}$, such that the length of tangent is equal to $k$ units itself. If particle starts form the point $(1,0)$, then

(a) the particle will cross $\mathrm{x}$-axis again at $\mathrm{x}=3$

(b) the particle will cross $x$-axis again at $x=4$

(c) the particle will cross +ve $x$-axis again at $x=2\sqrt{2}$

(d) the particle will cross +ve $x$-axis again at $x\in (2\sqrt{2},4)$

15. Let the particle starts from the point $(2,0)$ and moves $\pi /2$ units, on circle ${\mathrm{C}}_2$ in the counterclockwise direction, then itn moves on the circle ${\mathrm{C}}_3$ along the tangential path, let this straight line (tangential path traced by particle) intersect the circle ${\mathrm{C}}_3$ at the points. A and B tangents drawn at $\mathrm{A}$ and $\mathrm{B}$ intersect at

(a) $({\displaystyle \frac{9}{2\sqrt{2}}};{\displaystyle \frac{9}{2\sqrt{2}}})$

(b) $(9\sqrt{2},9\sqrt{2})$

(c) $(9,9)$

(d) $(\sqrt{2},\sqrt{2})$