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-2 y=3 . $
$\therefore-a+2 a=3 \Rightarrow a=$ $3 $
$\therefore$ Required equation of the circle $=(x+3)^2+(y+3)^2=9=x^2+y^2+6 x+6 y+9=0$ If the circle lies in the fourth quadrant, then its centre will be $(a,-a), $
$\therefore a+2 a=3 \Rightarrow a=1 $
$\therefore$ Required equation of the circle $=(x-1)^2+(y+1)^2=1=x^2+y^2-2 x+2 y+1=0$
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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) $\text{x}^2+\text{y}^2-\frac{29}{3}\times-\frac{19}{3}\text{y}+\frac{56}{3}=0$ $\Rightarrow3(\text{x}^2+\text{y}^2)-29\text{x}-19\text{y}+56=0$
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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 $x^2+y^2+2 g x+2 f y+c=0 \ldots \ldots .$. (1) We have, $P(0,0), Q(-2,1)$ and $R$ $(-3,2) P, Q \& R$ lies on(1), so, $0+00+0 g+c=0$............ (2) $4+1+4 x-2 y+c=0$ ............ (3) $9+4+6 x-4 y+c=0$
............ (4) Solving (2), (3) & (4), we get, $\mathrm{g}=-\frac{3}{2}, \mathrm{f}=-\frac{11}{2}=-\frac{11}{2}, \mathrm{c}=0$ from (1), Thus, equation of eirel e is, $\mathrm{x}^2+\mathrm{y}^2$ $+3 x-11 y=0$
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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 $2 x-y=3$. Let the equation of the circle be $x^2$ $+y^2+2 g x+2 f y+c=0$. Since the circle passes through $(3,-2)$ andAlso $(-2,0)$ therefore $9+4+6 g-4 f+c=0 . . . . . .$.
(1) $4+0-4 \mathrm{~g}+0+\mathrm{c}=0$.
(2) Also the centre of the circle lies on $2 x-y=3-2 g+f=3$
(3) Solving equallons
(1), (2) and (3), we get $g=\frac{3}{2}, f=6$ and $c=2$ Therefore the equation of the circle is $x^2+y^2+3 x+12 y+2=0$
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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, $(\text{x}\pm8)^2+(\text{y}+15)^2=17^2$ $\Rightarrow\text{x}^2+\text{y}^2\mp16\text{x}+30\text{y}=0$
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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 $(\text{x}-5)^2+(\text{y}-6)^2=6^2$ $\Rightarrow\text{x}^2+\text{y}^2-10\text{x}-12\text{y}+25=0$
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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 $x^2 + y^2 + 2gx + 2fy + c = 0$ .......... (1) centre = (-g, -f) and Rsdius $=\sqrt{\text{g}^2+\text{f}^2-\text{c}}$ $\therefore$ P = (5, 5), Q (6, 4) Be R = (-2, 4) lies on (1) $\therefore$ 25 + 25 + 10g + 10f + c = 0 ........ (2) 36 + 16 + 12g + 8f + c = 0 ............. (3) 4 + 16 + 4g + 8f + c = 0 ................ (4) Solving (2) (3) & (4), we get g = -2, f = -1 & c = -20 from (1) The equation of cirde is $x^2 + y^2+ 4x + 2y - 20 = 0$ Clearly S = (7, 1) Satisfy (A) Hence P,Q,R,S are concydic Now centre = (-g, -f) = (2, 1) radius $\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{4+1+20}=\sqrt{25}=5$
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 $(\text{x}-6)^2+(\text{y}-6)^2=6^2$ $\Rightarrow\text{x}^2+\text{y}^2-12\text{x}-12\text{y}+36=0$
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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~ 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, $(x - 1)^2+ (y + 2)^2 = 25 x^2 - 2x + 1 + y^2 + 4y + 4 = 25 x^2 + y^2 - 2x + 4y - 20 = 0$
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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 $(x-a)^2+(y-b)^2=r^2$. $\qquad$ (A) is $2 x+y=6$ $\qquad$ (1) $3 x+2 y=4$ $\qquad$ (2) The point of intersection of (1) \& (2) is $C=(8,-10)$, which is the centre of circle. Also, radius $=10 \therefore(A) \Rightarrow(x-8)^2+(y$ $+10)^2=10^2 \Rightarrow x^2+y^2-16 x+20 y+64=0$
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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, $(\text{x}-3)^2+(\text{y}-4)^2=\Big(\frac{62}{13}\Big)^2$ $\Rightarrow169[\text{x}^2+\text{y}^2-6\text{x}-8\text{y}]+25\times169=3844$ $\Rightarrow169[\text{x}^2+\text{y}^2-6\text{x}-8\text{y}]+381=0$
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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 $X-4 y=1$ .........(1) The genral equation of circle is $x^2+y^2+$ $2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$.........(2)
$\therefore P \& Q$ lies cn (2), so, $9+49+6 g+14 f+c=0$.........(3) $25+25+10 g+10 f+c=0$.........(4) Also, centre (-g, -f ) lies on (1)
$\therefore-\mathrm{g}+4 \mathrm{f}=1$.........(5) Solving (3) (3) (4) & (5) we get $\mathrm{g}=3, \mathrm{f}=1 \& \mathrm{c}=-90$ from
(2) The equation of cirde is $x^2+y^2+6 x+2 y-90=0$
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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 $x^2+y^2+2 g x+2 f y+c=0$ ........ (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, $25+64+10 g+16 f+c=0$ ........ (2) $4+81+4 g-18 f+c=0$
........ (3) $4+1+4 g-2 f+c=0$ ........ (4) Solving (2), (3) \& (4), we get, $g=58, f=24 & c=-285$ Thus, equation of eirel e is, $x^2+y^2+116 x-48 y-285=0$ from (1)
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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
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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, $(x - 1)^2 + (y + 1)^2 = 72 x^2 - 2x + 1 + y^2 + 2y + 1= 49 x^2 - 2x + y^2 + 2y = 47$
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