(Each question 2 marks)

Question 12 Marks

Find the equation of a circle, Which touches both the axes and passes through the point (2, 1).

Answer

The circle touches both the axis at $\mathrm{A}=(\mathrm{a}, 0)$ and $\mathrm{B}=(0, \mathrm{a})$ so, the centre of circle will be $(\mathrm{a}, \mathrm{a})$ and redius $=\mathrm{a}$. so, the equation of circle is $(x-a)^2+(y-a)^2=a^2$........ (A) Now, (A) Passes through P $(2,1) \therefore(2-a)^2+(1-a)^2=a^2 \Rightarrow 4$ $4 a+a^2+(1-a)^2+a^2=a^2 \Rightarrow 5-6 a+a^2=0 \Rightarrow(a-5)(a-1)=0 \Rightarrow a=5$ or 1 Thus the equation of circle will be $x^2-$ $10 x+y^2-10 y+25=0, x^2+y^2-2 x-2 y+1=0$

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Question 22 Marks

Find the equation of the circle passing through the point of intersection of the lines $x+3 y=0$ and $2 x-7 y=0$ and whose centre is the point of intersection of the lines $x+y+1=0$ and $x-2 y+4=0$.

Answer

The given equations of lines are $x+3 y=0$ .............(1) $2 x-7 y=0$ .............(2) $x+y=-1$ .............(3) $x-2 y=-4$.............(4)
The general equation of circle with centre $(a, b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$.............(A) Centre of ( $A$ ) is the point of intersection of (iii) & (iv)
$\therefore$ Centre $=(-2,1) \therefore(A) \Rightarrow(x+2)^2+(y-1)^2=r^2$,.............(B) Also, (A) passes through point of intersection of (1) & (2), that is through $P=(0,0) \therefore 2^2(-1)^2=r^2 \Rightarrow r=\sqrt{5}$
Thus, the equation of required circle is $(x+2)^2+(y-1)^2=5$ or, $x^2+y^2+4 x-2 y=0$

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Question 32 Marks

Find the centre and radius of the following circles: $\text{x}^2+\text{y}^2-4\text{x}+6\text{y}=5$

Answer

The general equation of orde is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2-\text{r}^2$ or $\text{x}^2+\text{y}^2-2\text{ax}-2\text{by}+\text{a}^2+\text{b}^2=\text{r}^2\ .....(\text{A})$ Where (a, b),s the centre and r be the radius of the circle. $\text{x}^2+\text{y}^2-4\text{x}+6\text{y}=5$ $\text{x}^2+\text{y}^2-4\text{x}+6\text{y}-5$ $\Rightarrow(\text{x}^2-4\text{x}+4)+(\text{y}^2+6\text{y}+9)=5+4+9$ $\Rightarrow(\text{x}-2)^2+(\text{y}+3)^2-(3\sqrt{2})^2$ Comparing with (A), we get centre = (2, -3) Radius $-3\sqrt{2}$

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Question 42 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 $2x + 5y = 18.$

Answer

Intersection 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 $=\sqrt{(1-4)^2+(-2-2)^2}$ $=\sqrt{9+16}$ $=5$ 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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Question 52 Marks

Find the coordinates of the centre and radius of each of the following circles:$\frac{1}{2}(\text{x}^2+\text{y}^2)+\text{x}\cos\theta+\text{y}\sin\theta-4=0$

Answer

The given equation can be rewritten as $\text{x}^2+\text{y}^2+2\text{x}\cos\theta+2\text{y}\sin\theta-8=0.$ $\therefore$ Centre $=(-\cos\theta,\ \sin\theta)$ And, radius $=\sqrt{(-\cos\theta)^2(-\sin\theta)^2+8}=\sqrt{1+8}=3$

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Question 62 Marks

Find the coordinates of the centre and radius of each of the following circles: $\text{x}^2 + \text{y}^2 + 6\text{x} − 8\text{y} − 24 = 0$

Answer

The given equation can be rewritten as $x^2 + y^2 + 2(3)x - 2(4)y - 24 = 0 \therefore$ Centre = (-3, -4) And, radius $=\sqrt{(3)^2+(4)^2+24}=\sqrt{49}=7$

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Question 72 Marks

Find the equation of the circle whose centre lies on the positive direction of y-axis at a distance 6 from the origin and whose radius is 4.

Answer

The genral equation of circle, with centre $(a, b)$ and radius $r$ in $(x-a)^2+(y-b)^2=r^2$............(A) Now, According to the question centre $=(0,6)$ and radius $=4 \therefore(A) \Rightarrow(x-0)^2+(y-6)^2=4^2 \Rightarrow x^2+y^2-12 y+20=0$

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Question 82 Marks

Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

Answer

We know that the equation of circle whose centre in (a, b) and radius r is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(1)$ We have centre = (1, 2) $\therefore(\text{x}-1)^2+(\text{y}-2)^2\ .....(2)$ Also, circle passes through ( 4, 6) $\therefore(4-1)^2=(6-2)^2=\text{r}^2$ $\Rightarrow9+16=\text{r}^2$ $\Rightarrow\text{r}=5$ Thus, equation of required circle in $(\text{x}-1)^2+(\text{y}-\text{2})^2=5^2$ or, $\text{x}^2+\text{y}^2-2\text{x}-4\text{y}-20=0$

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Question 92 Marks

Find the coordinates of the centre and radius of each of the following circles: $2\text{x}^2+2\text{y}^2-3\text{x}+5\text{y}=7$

Answer

The given equation can be rewritten as $\text{a}^2+\text{y}^2-\frac{3\text{x}}{2}+\frac{5\text{y}}{2}=\frac{7}{2}=0.$ $\therefore$ Centre $=\Big(\frac{3}{4},\ \frac{-5}{4}\Big)$ And, radius $=\sqrt{\Big(\frac{3}{4}\Big)^2+\Big(\frac{-5}{4}\Big)+\frac{7}{2}}=\sqrt{\frac{34+56}{16}}=\sqrt{\frac{90}{16}}=\frac{3\sqrt{10}}{4}$

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Question 102 Marks

Find the centre and radius of the following circles: $\text{x}^2+\text{y}^2-\text{x}+2\text{y}-3=0$

Answer

lhe general equation of orde is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2-\text{r}^2$ or $\text{x}^2+\text{y}^2-2\text{ax}-2\text{by}+\text{a}^2+\text{b}^2=\text{r}^2\ .....(\text{A})$ Where (a, b),s the centre and r be the radius of the circle. $\text{x}^2+\text{y}^2-\text{x}+2\text{y}=3$ $\Rightarrow\big(\text{x}^2-\text{x}+\frac{1}{4}\big)+(\text{y}^2+2\text{y}+1)=3+\frac{1}{4}+1$ $\Rightarrow\big(\text{x}-\frac{1}{2}\big)^2(\text{y}+1)^2=\Big(\frac{\sqrt{17}}{2}\Big)^2$ Companng with (A), we get centre $=\Big(\frac{1}{2},-1\Big)$ Radius $-\frac{\sqrt{17}}{2}$

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Question 112 Marks

Find the centre and radius of the following circles: $(\text{x}+5)^2+(\text{y}+1)^2=9$

Answer

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Question 122 Marks

Find the coordinates of the centre and radius of each of the following circles:$\text{x}^2+\text{y}^2-\text{ax}-\text{by}=0$

Answer

The given equation can be rewritten as $\text{x}^2+\text{y}^2=\frac{2\text{ax}}{2}-\frac{2\text{by}}{2}=0.$ $\therefore$ Centre $=\Big(\frac{\text{a}}{2},\ \frac{\text{b}}{2}\Big)$ And, radius $\sqrt{\Big(\frac{\text{a}}{2}\Big)^2+\Big(\frac{\text{b}}{2}\Big)^2}=\frac{1}{2}\sqrt{\text{a}^2+\text{b}^2}$

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MATHS (Each question 2 marks)

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