Question 13 Marks
Find the equation of the circle which touches the axes and whose centre lies on $x - 2y = 3.$
AnswerIf the circle lies in the third quadrant, then its centre will be $(-a, -a).$
The centre lies on $x - 2y = 3.$
$\therefore -a + 2a = 3 \Rightarrow a = 3$
$\therefore$ Required equation of the circle $= (x + 3)^2 + (y + 3)^2 = 9$
$= x^2 + y^2 + 6x + 6y + 9 = 0$
If the circle lies in the fourth quadrant, then its centre will be $(a, -a),$
$\therefore a + 2a = 3 \Rightarrow a = 1$
$\therefore$ Required equation of the circle $ = (x - 1)^2 + (y + 1)^2 = 1$
$= x^2 +y^2 − 2x + 2y + 1 = 0$
View full question & answer→Question 23 Marks
Find the equation of the circle passing through the points:
$(5, 7), (8, 1)$ and $(1, 3)$
AnswerWe know that the general equation of cirde is $x^2 + y^2+ 2gx + 2fy + c = 0 ......... (1)$
we have,
$P (5, 7), Q (8, 1)$ and $R (1, 3)$
Since $P, Q$ and $R$ lies on $(1)$
so,
$25 + 49 + 10g + 14f + C = 0 .......... (2)$
$64 + 1 + 16g + 2f +c = 0 ............... (3)$
$1 + 9 + 2g + 6f + c = 0 .................. (4)$
Solving $(2), (3)$ & $(4),$ we get,
$\text{g}=-\frac{29}{6},\ \text{f}=\frac{19}{6},\ \text{c}=\frac{56}{3}$
Thus, the equation of circle is on putting $g, f$ & con $(1)$
$\text{x}^2+\text{y}^2-\frac{29}{3}\times-\frac{19}{3}\text{y}+\frac{56}{3}=0$
$\Rightarrow3(\text{x}^2+\text{y}^2)-29\text{x}-19\text{y}+56=0$
View full question & answer→Question 33 Marks
Find the equation of the circle passing through the points:
$(0, 0), (-2, 1)$ and $(-3, 2)$
AnswerWe know that the general equation of circle is $x^2 + y^2 + 2gx + 2fy + c = 0 ........ (1)$
We have,
$P (0, 0), Q (-2, 1)$ and $R (-3, 2)$
$P, Q$ & $R$ lies on$(1)$, so,
$0 + 00 + 0g + c = 0 ............ (2)$
$4 + 1 + 4x - 2y + c = 0 ........ (3)$
$9 + 4 + 6x - 4y + c = 0 ........ (4)$
Solving (2), (3) & (4), we get,
$\text{g}=-\frac{3}{2},\ \text{f}=-\frac{11}{2}=-\frac{11}{2},\ \text{c}=0$
from $(1),$
Thus, equation of eirel e is,
$x^2 + y^2 + 3x - 11y = 0$
View full question & answer→Question 43 Marks
Find the equation of the circle which passes through $(3, -2), (-2, 0)$ and has its centre on the line $2x - y = 3$
AnswerA circle pass Ing through $P(3, -2)$ and $Q(-2, 0) $ and having its centre on $2x - y = 3.$
Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0.$
Since the circle passes through $(3, -2)$ and Also $(-2, 0)$ therefore
$9 + 4 + 6g - 4f + c = 0 ......... (1)$
$4 + 0 - 4g + 0 + c = 0 .......... (2)$
Also the centre of the circle lies on $2x - y = 3$
$-2g + f = 3 ....... (3)$
Solving equallons $(1), (2)$ and $(3),$ we get
$\text{g}=\frac{3}{2},\ \text{f}=6$ and $\text{c}=2$
$Therefore the equation of the circle Is$
$x^2 + y^2 + 3x + 12y + 2 = 0$
View full question & answer→Question 53 Marks
Find the equation of the circle passing through the points:
$(1, 2), (3, -4)$ and $(5, -6)$
AnswerWe know that the general equation of circle is $x^2 + y^2 + 2gx + 2fy + c = 0 ........ (1)$
We have,
$P (1,2), Q (3, -4)$ and $R (5, -6)$
Since $P, Q$ & $R$ lies on $(1)$
$\therefore x^2 + y^2 + 2gx + 2fy + c = 0 ...... (1)$
$1 + 4 + 2g + 4f + c = 0 ................ (2)$
$9 + 16 + 6g - 8f + c = 0 ............... (3)$
$25 + 36 + 10g - 12f + c = 0 .......... (4)$
Solving $(2), (3)$ & $(4),$ we get,
$g = -11, f = -2$ & $c = 25$
from $(1)$
The equation of circle is
$x^2 + y^2 -22x - 4y + 25 = 0$
View full question & answer→Question 63 Marks
Find the equation of a circle,
Passing through the origin, radius 17 and ordinate of the centre is -15.
AnswerThe circle passes through origin (0, 0) and has radius = 17 units
Also, the ordinate of centre is -15 then assume abssisa is a.
$\therefore\text{OC}=17$
$\Rightarrow\sqrt{(\text{a}-0)^2+(0+15)^2}=17$ (by distance form ulla)
$\Rightarrow\sqrt{\text{a}^2+225}=17$
$\Rightarrow\text{a}^2+225=289$
$\Rightarrow\text{a}^2=64$
$\Rightarrow\text{a}-\pm8$
$\therefore$ Centre $=(\pm8,\ -15)$
Thus, the equation of circle will be,
$(\text{x}\pm8)^2+(\text{y}+15)^2=17^2$
$\Rightarrow\text{x}^2+\text{y}^2\mp16\text{x}+30\text{y}=0$
View full question & answer→Question 73 Marks
Find the equation of a circle,
Which touches x-axis at a distance 5 from the origin and radius 6 units.
AnswerThe circle touches the x-axis at A = (5, 0) and has radius 6 unit
Thus,
centre = (5, b)
By distance form ulla OA = 6
$\Rightarrow\sqrt{(5-5)^2+(\text{b}-0)^2}=6$
$\Rightarrow\text{b}=6$
$\Rightarrow$ Centre $=(5,\ 6)$
so, the equation of required cirde is
$(\text{x}-5)^2+(\text{y}-6)^2=6^2$
$\Rightarrow\text{x}^2+\text{y}^2-10\text{x}-12\text{y}+25=0$
View full question & answer→Question 83 Marks
Find the equation of the circle which circumscribes the triangle formed by the lines
$x + y + 3 = 0, x - y + 1 = 0$ and $x = 3$
AnswerThe given equation of lines
$x + y = -3 ......... (1)$
$x - y = -1 ......... (2)$
$x = 3 .............. (3)$
Let A, B, C are the point of intersection of lines (1) & (2), (2) & (3) and (3) & (1) respectively
$\therefore$ $A = (-2, -1), B = (3, 4)$ & $C = (3, -6)$
Now A cirde $x^2 + y^2+ 2gx + 2fy + c = c ........ (A)$
circumscribing the $\Delta\text{ABC}$
$\therefore$ $4 + 1 - 4g - 2f + c = 0 ........ (4)$
$9 + 16 + 6g + af + c = 0 ........ (5)$
$9 + 36 + 6g - 12f + c = 0 ....... (6)$
Solving (4), (5) & (6) we get,
$g = -3, f = 1, c = -15$
from (A),
The equation of required cirde is
$x^2 + y^2 - 6x + 2y - 15 = 0$
View full question & answer→Question 93 Marks
Show that the points $(5, 5), (6, 4), (-2, 4)$ and $(7, 1)$ all lie on a circle, and find its equation, centre and radius.
AnswerThe genral equation of circle is
$x^2+y^2+2 g x+2 f y+c=0$
$\text { centre }=(-g,-f) \text { and }$
$\text { Rsdius }=\sqrt{g^2+f^2-c}$
$\therefore P=(5,5), Q(6,4) Be R=(-2,4) \text { lies on (1) }$
$\therefore 25+25+10 g+10 f+c=0$
$36+16+12 g+8 f+c=0$
$4+16+4 g+8 f+c=0$
Solving (2) (3) & (4), we get
$g=-2, f=-1 \& c=-20$
from (1)
The equation of cirde is
$x^2+y^2+4 x+2 y-20=0$
Clearly $S=(7,1)$ Satisfy $(A)$
Hence $P,Q,R,S$ are concydic
Now
centre $=(-g,-f)=(2,1)$
radius $\sqrt{ g ^2+ f ^2- c }=\sqrt{4+1+20}=\sqrt{25}=5$
View full question & answer→Question 103 Marks
Find the equation of the circle which circumscribes the triangle formed by the lines
$2x + y - 3 = 0, x + y - 1 = 0$ and $3x + 2y - 5 = 0$
AnswerThe given equation of lines
$2x + y = 3 ....... (1)$
$x - y = 1 .......... (2)$
$3x + 2y = 5 ......... (3)$
Let $A, B$ & $C$ are the point of intersection of lines (1) & (2), (2) & (3) and (3) & (1) respectively
$\therefore$ $A = (2, -1), B = (3, -2)$ & $C = (1, 1)$
Let $x^2 + y^2+ 2gx + 2fy + c = 0 ........ (A)$
be the circle that circum scnbinq $\Delta\text{ABC}$
$\therefore$ $4 + 1 - 4g - 2f + c = 0 ........ (4)$
$9 + 16 + 6g + 2f + c = 0 ........ (5)$
$1 + 1 + 2g + 2f + c = 0 ......... (6)$
Solving (4), (5) & (6) we get,
$\text{g}=-\frac{13}{2},\ \text{f}=\frac{5}{2}\ \&\ \text{c}=16$
from (A),
The required cirde is
$x^2 + y^2 - 13x + 5y - 16 = 0$
View full question & answer→Question 113 Marks
Find the equation of a circle,
Which touches both the axes at a distance of 6 units from the origin.
AnswerThe circle touches the axes at (0, 6) and (6, 0) respectively
Thus, the centre of circle will be (6, 6) ( as shown in fig)
and radius $=\text{OA}=\sqrt{(6-0)^2+(6-6)^2}=\sqrt{36}=6$ (by distance formulla)
$\therefore$ The equation of circle will be $(\text{x}-6)^2+(\text{y}-6)^2=6^2$
$\Rightarrow\text{x}^2+\text{y}^2-12\text{x}-12\text{y}+36=0$
View full question & answer→Question 123 Marks
Find the equation of the circle having $(1, -2)$ as its centre and passing through the intersection of the lines $3x + y = 14$ and $2x + 5y = 18.$
AnswerIntersection of $3x + y = 14$ and $2x +5y= 18$ is
Obtained by solving two equations.
$x = 4 $ and $y = 2$
Point $(4,2)$ is on circle, hence$\sim$ distance from centre
$(1,-2)$
= Radius
$=\sqrt{(1-4)^2+(-2-1)^2}$
$=\sqrt{9+16}$
$=5$
Equation of the circle with centre $(4,2)$ and radius $5$ is,
$(x - 1)^2+ (y + 2)^2 = 25$
$x^2 - 2x + 1 + y^2 + 4y + 4 = 25$
$x^2 + y^2 - 2x + 4y - 20 = 0$
View full question & answer→Question 133 Marks
If the equations of two diameters of a circle are $2x + y = 6$ and $3x + 2y = 4$ and the radius is $10,$ find the equation of the circle.
AnswerThe equation of two diameters of the circle $(x - a)^2 + (y - b)^2 = r^2 ......... (A)$
$is 2x + y = 6 ......... (1)$
$3x + 2y = 4 .......... (2)$
The point of intersection of $(1)$ & $(2)$ is $C = (8, -10),$ which is the centre of circle.
Also, radius $= 10$
$\therefore$ (A)
$\Rightarrow (x - 8)^2+(y + 10)^2= 10^2$
$\Rightarrow x^2 + y^2- 16x + 20y + 64 = 0$
View full question & answer→Question 143 Marks
Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y − 1 = 0.
AnswerThe centre of the required circle in (3,4) and the circle touches the line 5x + 12y = 1
so, radius = OA = Perpendicular distance of O to 5x + 12y = 1
$[\therefore$ radius is perpendicular to the tangent$]$
$\Rightarrow\text{OA}\frac{5\times3+12\times4-1}{\sqrt{5}^2+12^2}$
$=\frac{62}{13}$
Thus the equation of circle will be,
$(\text{x}-3)^2+(\text{y}-4)^2=\Big(\frac{62}{13}\Big)^2$
$\Rightarrow169[\text{x}^2+\text{y}^2-6\text{x}-8\text{y}]+25\times169=3844$
$\Rightarrow169[\text{x}^2+\text{y}^2-6\text{x}-8\text{y}]+381=0$
View full question & answer→Question 153 Marks
Find the equation of the circle which passes through the points $(3, 7), (5, 5)$ and has its centre on the line $x - 4y = 1.$
AnswerThe circle passes through $P$ & $Q$ and the centre lies on
$X - 4y = 1 .........(1)$
The genral equation of circle is
$x^2 + y^2 + 2gx + 2fy + c = 0 .......... (2)$
$\therefore P$ & $Q$ lies cn $(2),$ so,
$9 + 49 + 6g + 14f + c = 0 ............ (3)$
$25 + 25 + 10g + 10f + c = 0 ......... (4)$
Also, centre $(-g, -f)$ lies on $(1)$
$\therefore - g + 4f = 1 .......... (5)$
Solving $(3) (4)$ & $(5)$ we get
$g = 3, f = 1$ & $c = -90$
from $(2)$
The equation of cirde is
$x^2 + y^2 + 6x + 2y - 90 = 0$
View full question & answer→Question 163 Marks
Find the equation of the circle passing through the points:
$(5, -8), (-2, 9)$ and $(2, 1)$
AnswerWe know that the general equation of circle is $x^2 + y^2 + 2gx + 2fy + c = 0 ........ (1)$
We have,
$P (5, -8), Q (-2, 9)$ and $R (2, 1)$
Since $P, Q$ & $R$ lies on $(1)$
$P, Q$ & $R$ lies on$(1)$, so,
$25 + 64 + 10g + 16f + c = 0 ........ (2)$
$4 + 81 + 4g - 18f + c = 0 ............ (3)$
$4 + 1 + 4g - 2f + c = 0 ................ (4)$
Solving (2), (3) & (4), we get,
$g = 58, f = 24$ & $c = -285$
Thus, equation of eirel e is,
$x^2 + y^2 + 116x - 48y - 285 = 0$ from (1)
View full question & answer→Question 173 Marks
Show that the points $(3, -2), (1, 0), (-1, -2)$ and $(1, -4)$ are concyclic.
Answerwe have,
$P = (3, -2), Q = (1,0), R = (-1,-2)$ and $S = (1, -4)$
let us consider $A$ circle $x^2 + y^2 + 2gx + 2fy + c = 0 ........ (1)$
Passes through $P, Q$ & $R$
$\therefore 9 + 4 + 6g - 4f + c = 0 ........ (2)$
$1 + 0 + 2g - 0 + c = 0 ............ (3)$
$1 + 4 - 2g - 4f + c = 0 ............ (4)$
Solving $(2), (3)$ & $(4)$ we get,
$g = -1, f = 2$ & $c = 1$
from$(1)$
The required equation of ercle is
$x^2 + y^2 - 2x + 4y + 1 = 0 ......... (5)$
Clearly $s = (1, -4)$ satisfy $(5)$
Thus,
$P, Q, R$ & $S$ are concydic
View full question & answer→Question 183 Marks
If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154$ square units, then obtain the equation of the circle.
AnswerArea of given circle is $= 154$
$\pi^2=154$
$\frac{22}{7}\text{r}^2154$
$\text{r}^2=154\times\frac{7}{22}$
$\text{r}^2=154\times\frac{7}{22}$
$\text{r}^2=49$
$\text{r}=7$
The intersection point of $2x - 3y = 5$ and $3x - 4y = 7$ is
The centre of the circle.
Solving simultaneous equations
$2x - 3y = 5$ and $3x - 4y = 7$ we get,
Centre of circle as $(1, -1)$
Equation of circle with centre $(1, -1)$ and radius$= 7$ is,
$(x - 1)^2 + (y + 1)^2 = 72$
$x^2 - 2x + 1 + y^2 + 2y + 1= 49$
$x^2 - 2x + y^2 + 2y = 47$
View full question & answer→