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MATHS 2 Marks Questions

The Circle · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 12 questions.

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Question 12 Marks
Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x - 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x - 2y + 4 = 0.
Answer
The given equations of lines are
x + 3y = 0 .............(1)
2x - 7 y = 0 ...........(2)
x + y = -1 ..............(3)
x - 2y = -4 ........... (4)
The general equation of circle with centre (a, b) and radius r is
(x - a)2 + (y - b)2 = r2 ...........(A)
Centre of ( A) is the point of intersection of (iii) & (iv)
$\therefore$ Centre = (-2, 1)
$\therefore$ (A)
⇒ (x + 2)2 + (y - 1)2 = r2 ......... (B)
Also, (A) passes through point of intersection of (1) & (2), that is through P = (0, 0)
$\therefore2^2(-1)^2=\text{r}^2\Rightarrow\text{r}=\sqrt{5}$
Thus, the equation of required circle is
(x + 2)2 + (y - 1)2 = 5
or,
x2 + y2 + 4X - 2y = 0
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Question 22 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 32 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
x2 - 2x+ 1 + y2 + 4y + 4 = 25
x2 + y2 - 2x + 4y - 20 = 0
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Question 42 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 52 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 x2 + y2 + 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 62 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 = r2 ............ (A)
Now,
According to the question
centre = ( 0, 6) and radius = 4
$\therefore$ (A)
⇒ (x - 0)2 + (y - 6)2 = 42
⇒ x2 + y2 - 12y + 20 = 0
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Question 72 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 82 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 92 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 102 Marks
Find the centre and radius of the following circles:
$(\text{x}+5)^2+(\text{y}+1)^2=9$
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.
$(\text{x}+5)^2+(\text{y}+1)^2=9$
$\Rightarrow(\text{x}+5)^2+(\text{y}+1)^2=3^2$
comparing with (A) we get,
Centre = ( -5, -1)
Radius -3
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Question 112 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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Question 122 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 A = (a, 0) and B = (0, a)
so, the centre of circle will be (a, a) and redius = a.
so, the equation of circle is (x - a)2 + (y - a)2 = a2 ............ (A)
Now,
(A) Passes through P (2, 1)
$\therefore$ (2 - a)2 + (1 - a)2 = a2
⇒ 4 - 4a + a2 + (1 - a)2 + a2 = a2
⇒ 5 - 6a + a2 = 0
⇒ (a - 5)(a - 1) = 0
⇒ a = 5 or 1
Thus the equation of circle will be
 x2 - 10x + y2 - 10y + 25 = 0, x2 + y2 - 2x - 2y + 1 = 0
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