Coordinate Geometry-i Circles (Lecture-02)
1. Find the equation of the system of circles co-axial with the circles $x^{2}+y^{2}+4 x+2 y+1=0$ and $x^{2}+y^{2}-$ $2 x+6 y-6=0$. Also find the equation of that particular circles whose centre lies on radical axis.
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Solution:
Given circles are
$\begin{aligned} & S _{1} \equiv x^{2}+y^{2}+4 x+2 y+1=0 \\ & S _{2} \equiv x^{2}+y^{2}-2 x+6 y-6=0 \\ & S _{1}-S _{2}=0 \\ & 6 x-4 y+7=0 \end{aligned}$
System of co-axial circle is $\mathrm{S} _{1}+\lambda\left(\mathrm{S} _{1}-\mathrm{S} _{2}\right)=0$
$\begin{aligned} & x^{2}+y^{2}+4 x+2 y+1+\lambda(6 x-4 y+7)=0 \\ & x^{2}+y^{2}+2 x(2+3 \lambda)+2 y(1-2 \lambda)+1+7 \lambda=0 \end{aligned}$
Centre of this circle is $(-(2+3 \lambda),-(1-2 \lambda)$
lies on radical axis
$\begin{aligned} & \therefore 6(-2-3 \lambda)+4(1-2 \lambda)+7=0 \\ & -12-18 \lambda+4-8 \lambda+7=0 \\ & -1-26 \lambda=0 \\ & \lambda=\dfrac{-1}{26} \end{aligned}$
$\therefore$ Required particular member of co-axial circle is $26\left(x^{2}+y^{2}\right)+98 x+56 y+19=0$
2. If the circumference of the circle $x^{2}+y^{2}+8 x+8 y-b=0$ is bisected by the circle $x^{2}+y^{2}-2 x+4 y+a=0$ then $a+b$ equal to
(a) 50
(b) 56
(c) -56
(d) -34
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Solution: (c)
Equation of radical axis (common chord of these circles) is $10 x+4 y-b-a=0$
Centre of first circle is $(-4,-4)$
Since second circle bisects the first circle
Therefore centre of first circle must lie on common chord.
$\begin{array}{ll} \therefore \quad & 10(-4)+4(-4)-b-a=0 \\ & -40-16-(a+b)=0 \\ \therefore & a+b=-56 \end{array}$
3. The equation of the circle passing through the point of intersection of the circles $x^{2}+y^{2}-4 x-2 y=8$ and $x^{2}+y^{2}-2 x-4 y=8$ and the point $(-1,4)$ is
(a) $x^{2}+y^{2}+4 x+4 y-8=0$
(b) $x^{2}+y^{2}-3 x+4 y+8=0$
(c) $x^{2}+y^{2}+x+y-8=0$
(d) $x^{2}+y^{2}-3 x-3 y-8=0$
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Solution: (d)
Equation of any circle passing through the point of intersection of the circles is $\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-2 \mathrm{y}-8+\lambda\left(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}-8\right)=0$
This circle passes through the point $(-1,4)$
$\begin{array}{ll} \therefore \quad & 1+16+4-8-8+\lambda(1+16+2-16-8)=0 \\ & 5-5 \lambda=0 \\ & \lambda=1 \end{array}$
Required circle is $\mathrm{x}^{2}+\mathrm{y}^{2}-3 \mathrm{x}-3 \mathrm{y}-8=0$
4. If the common chord of the circles $x^{2}+(y-b)^{2}=16$ and $x^{2}+y^{2}=16$ subtends a right angle at the origin then $b=$
(a) 4
(b) $4 \sqrt{2}$
(c) $-4 \sqrt{2}$
(d) 8
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Solution:
The equation of common chord is
$\begin{array}{ll} & S-S _{1}=0 \\ & (y-b)^{2}-y^{2}=0 \\ & b^{2}-2 b y=0 \\ & b(b-2 y)=0 \\ \therefore \quad & b=2 y \text { or } 1=\dfrac{2 y}{b} \end{array}$
The combined equation of the straight lines joining the origin to the points of intersection $y=b / 2$
and $x^{2}+y^{2}=16\left(\dfrac{2 y}{b}\right)^{2} \Rightarrow b^{2} x^{2}+\left(b^{2}-64\right) y^{2}=0$
This equation represents a pair of perpendicular lines
$\therefore \quad b^{2}+b^{2}-64=0 \Rightarrow b= \pm 4 \sqrt{2}$
5. Given the circles $x^{2}+y^{2}-4 x-5=0$ and $x^{2}+y^{2}+6 x-2 y+6=0$ Let $P$ be a point $(\alpha, \beta)$ such that the tangents from $\mathrm{P}$ to both the circles are equal. Then
(a) $2 \alpha+10 \beta+11=0$
(b) $2 \alpha-10 \beta+11=0$
(c) $10 \alpha-2 \beta+11=0$
(d) $10 \alpha+2 \beta+11=0$
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Solution:
$\mathrm{PT} _{1}=\mathrm{PT} _{2}$
$\sqrt{\alpha^{2}+\beta^{2}-4 \alpha-5}=\sqrt{\alpha^{2}+\beta^{2}+6 \alpha-2 \beta+6}$
Squaring,
$\alpha^{2}+\beta^{2}-4 \alpha-5=\alpha^{2}+\beta^{2}+6 \alpha-2 \beta+6$
$10 \alpha-2 \beta+11=0$
correct option is ‘c’
6. If the circles $x^{2}+y^{2}+2 x+2 k y+6=0$ and $x^{2}+y^{2}+2 k y+k=0$ intersect orthogonaly then $k$ is
(a) 2 or $-3 / 2$
(b) $-2 \mathrm{or}-3 / 2$
(c) 2 or $3 / 2$
(d) $-2 \mathrm{or}3 / 2$
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Solution:
Condition for two circles to intersect at right angles is $2 \mathrm{~g} _{1} \mathrm{~g} _{2}+2 \mathrm{f} _{1} \mathrm{f} _{2}=\mathrm{c} _{1}+\mathrm{c} _{2}$
Here two circles are $x^{2}+y^{2}+2 x+2 k y+6=0$ and $x^{2}+y^{2}+2 k y+k=0$
$\mathrm{g} _{1}=1, \mathrm{f} _{1}=\mathrm{kc} _{1}=6$
$\mathrm{g} _{2}=0 \mathrm{f} _{2}=\mathrm{kc} _{2}=\mathrm{k}$
$\begin{array}{ll} \therefore \quad & 0+2 \mathrm{k}^{2}=6+\mathrm{k} \\ & 2 \mathrm{k}^{2}-\mathrm{k}-6=0 \\ & 2 \mathrm{k}^{2}-4 \mathrm{k}+3 \mathrm{k}-6=0 \\ & (2 \mathrm{k}+3)(\mathrm{k}-2)=0 \\ & \mathrm{k}=-3 / 2 \text { or } \mathrm{k}=2 \end{array}$
correct option is ‘a’
7. The distance between the chords of contact of the tangents to the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ from the origin and the point $(\mathrm{g}, \mathrm{f})$ is
(a) $\mathrm{g}^{2}+\mathrm{f}^{2}$
(b) $\dfrac{1}{2}\left(\mathrm{~g}^{2}+\mathrm{f}^{2}+\mathrm{c}\right)$
(c) $\dfrac{1}{2}\left(\dfrac{\mathrm{g}^{2}+\mathrm{f}^{2}+\mathrm{c}}{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}}}\right)$
(d) $\dfrac{1}{2} \dfrac{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}}}$
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Solution:
Equation of chord of contact to the circle from $\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ is
$\mathrm{xx} _{1}+\mathrm{yy} _{1}+\mathrm{g}\left(\mathrm{x}+\mathrm{x} _{1}\right)+\mathrm{f}\left(\mathrm{y}+\mathrm{y} _{1}\right)+\mathrm{c}=0$
From $(0,0)$ $\quad \quad \quad$ is $\mathrm{gx}+\mathrm{fy}+\mathrm{c}=0$……………..(1)
& From $(\mathrm{g}, \mathrm{f})$ $\quad \quad \quad$ is $\mathrm{gx}+\mathrm{fy}+\mathrm{g}(\mathrm{x}+\mathrm{g})+\mathrm{f}(\mathrm{y}+\mathrm{f})+\mathrm{c}=0$
$\hspace {3 cm}\begin{aligned} 2g x+2 f y+g^{2}+f^{2}+c=0 \end{aligned}$
$\hspace {3 cm}\begin{aligned} g x+f y+\left(\dfrac{g^{2}+f^{2}+c}{2}\right)=0……………..(2) \end{aligned}$
Now lines (1) & (2) are parallel
$\therefore \quad$ distance between paralled line is $\left|\dfrac{\mathrm{c} _{1}-\mathrm{c} _{2}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|$
$\therefore \quad\left|\dfrac{-\mathrm{c}+\dfrac{\mathrm{g}^{2}+\mathrm{f}^{2}+\mathrm{c}}{2}}{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}}}\right|=\dfrac{\dfrac{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}{2}}{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}}}=\dfrac{1}{2} \dfrac{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}}}$
correct option is ’d’
EXERCISE:
1. Let $0<\alpha<\dfrac{\pi}{2}$ be a fixed angle. If $\mathrm{P}=(\cos \theta, \sin \theta)$ and $\mathrm{Q}=\cos (\alpha-\theta), \sin (\alpha-\theta) \mathrm{Q}$ obtained form $\mathrm{P}$ by
(a) clockwise rotation around origin through an angle $\alpha$
(b) anti-clockwise rotation around origin through an angle $\alpha$
(c) reflection in the line through origin with slope $\tan \alpha$
(d) reflection in the line through origin with slope $\tan \alpha / 2$
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Answer: d2. If the tangent at the point $P$ on the circle $x^{2}+y^{2}+6 x+6 y=2$ meets the straight line $5 x-2 y+6=0$ at a point $\mathrm{Q}$ on the $\mathrm{y}$-axis, then the length of $\mathrm{PQ}$ is
(a) 4
(b) $2 \sqrt{5}$
(c) 5
(d) $3 \sqrt{5}$
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Answer: c3. The equations to the sides $\mathrm{AB}, \mathrm{BC}, \mathrm{CA}$ of a $\triangle \mathrm{ABC}$ are drawn on $\mathrm{AB}, \mathrm{BC}, \mathrm{CA}$ as diameters. The point of concurrence of the common chord is
(a) centroid of the triangle
(b) orthocenter
(c) circumcentre
(d) incentre
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Answer: b4. The number of rational points ( $a$ point $(a, b)$ is rational, if $a$ and $b$ both are rational numbers) on the circumference of a circle having centre $(\pi, \mathrm{e})$ is
(a) at most one
(b) at least two
(c) exactly two
(d) infinite
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Answer: a5. The locus of a point such that the tangents drawn from it to the circle $x^{2}+y^{2}-6 x-8 y=0$ are perpendicular to each other is
(a) $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}-8 \mathrm{y}-25=0$
(b) $x^{2}+y^{2}+6 x-8 y-5=0$
(c) $x^{2}+y^{2}-6 x+8 y-5=0$
(d) $x^{2}+y^{2}-6 x-8 y+25=0$
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Answer: a6. If the two circles $x^{2}+y^{2}+2 g x+2 f y=0$ and $x^{2}+y^{2}+2 g _{1} x+2 f _{1} y=0$ touch each other, then
(a) $f _{1} g=f g _{1}$
(b) $\mathrm{ff} _{1}=\mathrm{gg} _{1}$
(c) $\mathrm{f}^{2}+\mathrm{g}^{2}=\mathrm{f} _{1}^{2}+\mathrm{g} _{1}^{2}$
(d) none of these
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Answer: a7. The number of integral values of $\lambda$ for which $x 2+y 2+\lambda x+(1-\lambda) y+5=0$ is the equation of a circle whose radius cannot exceed 5 , is
(a) 14
(b) 18
(c) 16
(d) none of these
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Answer: c8. The circle $x^{2}+y^{2}+4 x-7 y+12=0$ cuts an intercept on $y$-axis of length
(a) 3
(b) 4
(c) 7
(d) 1
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Answer: d9. One of the diameter of the circle $x^{2}+y^{2}-12 x+4 y+6=0$ is given by
(a) $x+y=0$
(b) $x+3 y=0$
(c) $\mathrm{x}=\mathrm{y}$
(d) $3 x+2 y=0$
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Answer: b10. The coordinates of the middle point of the chord cut off by $2 x-5 y+18=0$ by the circle $x^{2}+y^{2}-6 x+2 y-54=0$ are
(a) $(1,4)$
(b) $(2,4)$
(c) $(4,1)$
(d) $(1,1)$
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Answer: aPASSAGE - 1
Let $A \equiv(a, 0)$ and $B \equiv(-a, 0)$ be two fixed points $\forall a \in(-\infty, 0)$ and $P$ moves on a plane such that $\mathrm{PA}=\mathrm{nPB}(\mathrm{n} \neq 0)$.
On the basis of above information, answer the following questions:
11. If $|\mathrm{n}| \neq 1$, then the locus of a point $\mathrm{P}$ is
(a) a straight line
(b) a circle
(c) a parabola
(d) an ellipse
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Answer: b12. If $\mathrm{n}=1$, then the locus of a point $\mathrm{P}$ is
(a) a straight line
(c) a circle
(c) a parabola
(d) a hyperbola
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Answer: a13. If $0<\mathrm{n}<1$, then
(a) A lies inside the circle and $\mathrm{B}$ lies outside the circle
(b) A lies outside the circle and $\mathrm{B}$ lies inside the circle
(c) both $\mathrm{A}$ and $\mathrm{B}$ lies on the circle
(d) both $A$ and $B$ lies inside the circle
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Answer: a14. If $\mathrm{n}>1$, then
(a) A lies outside the circle and $\mathrm{B}$ lies inside the circle
(b) A lies outside the circle and $B$ lies inside the circle
(c) both $\mathrm{A}$ and $\mathrm{B}$ lies on the circle
(d) both $\mathrm{A}$ and $\mathrm{B}$ lies inside the circle
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Answer: b15. If focus of $\mathrm{P}$ is a circle, then the circle
(a) passes through $\mathrm{A}$ and $\mathrm{B}$
(b) never passes through $A$ and $B$
(c) passes through $A$ but does not pass through $B$
(d) passes through $\mathrm{B}$ but does not pass through $\mathrm{A}$