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Maths 1 Marks Question

The Circle · Jharkhand Board (JAC) STD 11 Science (Jharkhand - English Medium). 15 questions.

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Question 11 Mark
If the line $y = mx$ does not intersect the circle $(x + 10)^2 + (y + 10)^2 = 180,$ then write the set of values taken by m.
Answer
$y = mx$ does not intersect the circle $(x + 10)^2 + (y + 10)^2 = 180$
Now, $(x + 10)^2+ (mx + 10)^2 = 180$
$x^2 +100 + 20x + m^2x^2 + 100 + 20mx = 180$
$x^2 + m^2x^2 + 20mx + 20x + 20 = 0$
$x^2 (1 + m^2) + 20x (m + 1) + 20 = 0$
Since $y = mx$ does not intersect so
$b^2 - 4ac < 0$
$[20(m + 1)^2 ] -4(1 + m^2)(20)< 0$
$400 (m + 1)^2 - 80 (1 + m^2) < 0$
$5m^2 + 5 + 10m - 1 - m^2 < 0$
$4m^2 + 10m + 4 < 0$
$4m^2 + 8m + 2m + 4 < 0$
$4m(m + 2) + 2 (m + 2) < 0$
$(m + 2)(4m + 2) < 0$
$\Rightarrow m + 2 < 0$ and $4m + 2 > 0$
$\Rightarrow m < -2$ and $m > -2$
Its not possible
$\Rightarrow\text{m}+2>0$ and $4\text{m}+2<0$
$\Rightarrow\text{m}>-2$ and $\text{m}<-\frac{1}{2}$
$\Rightarrow-2<\text{m}<-\frac{1}{2}$
$\Rightarrow\text{m}\in\Big(-2,\ \frac{1}{2}\Big)$
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Question 21 Mark
Write the area of the circle passing through (-2, 6) and having its centre at (1, 2).
Answer
Radius of circle = Distance between centre (1, 2) and (-2, 6)
$=\sqrt{(1+2)^2+(2-6)^2}$
$=\sqrt{9+16}$
$=5$
Area of the circle = $25\pi$
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Question 31 Mark
Write the coordinates of the centre of the circle inscribed in the square formed by the lines x = 2, x = 6, y = 5 and y = 9.
Answer
Cir de is inscribed in the square formed by lines x = 2, x = 6 and y = 5, y = 9
It is a square of side 4
Thus, abscissae of centre $=\frac{2+6}{2}=4$
Ordinate of centre $=\frac{5+9}{2}=7$
Centre of the circle = (4, 7)
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Question 41 Mark
Find the equation of the circle with:
Centre (a, b) and radius $\sqrt{\text{a}^2+\text{b}^2}$
Answer
The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b)are centre and r is radius
From (A)
$(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\big(\sqrt{\text{a}^2+\text{b}}\big)^2$
$\Rightarrow\text{x}^2-2\text{ax}+\text{a}^2+\text{y}^2-2\text{bx}+\text{b}^2=\text{a}^2+\text{b}^2$
$\Rightarrow\text{x}^2+\text{y}^2-2\text{ax}-2\text{by}=0$
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Question 51 Mark
If the radius of the circle $x^2 + y^2 + ax + (1 - a) y + 5 = 0$ does not exceed $5$, write the number of integral values a.
Answer
Equation of given circle is $x^2 + y^2 + ax + (1 - a)y + 5 = 0$
$\text{g}=\frac{\text{a}}{\text{2}},\ \text{f}=\frac{(1-\text{a})}{2}$
Radius $=\text{r}=\sqrt{\text{g}^2+\text{f}^2-\text{c}}$
$=\sqrt{\frac{\text{a}^2}{4}}+\frac{(1-\text{a}^2)}{4}-5<5$ .........[Radius can be at most]
$\frac{\text{a}^2}{4}+\frac{(1-\text{a})^2}{4}-5<25$
$\text{a}^2+(1-\text{a}^2)<120$
Sum of two squares should be I ess than $120.$
a can be at most $8$ and atleast $-7$.
So a can take $16$ integral values which are from $-7$ to $8$
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Question 61 Mark
Write the length of the intercept made by the circle $x^2 + y^2 + 2x - 4y - 5 = 0$ on y-axis.
Answer
$\text{x}^2+\text{y}^2+2\text{x}-4\text{y}-5=0$
Put $x = 0$
$\text{y}^2-4\text{y}-5=0$
$\text{y}^2-5\text{y}+\text{y}-5=0$
$\text{y}(\text{y}-5)+(\text{y}-5)=0$
$(\text{y}-5)(\text{y}+1)=0$
$\text{y}=5,-1$
Thus, circle cuts y-axis at $(0, 5)$ and $(0, -1)$ so
length of intercept on y-axisby circle $= 5 + 1 = 6$ units.
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Question 71 Mark
Find the equation of the circle with:
Centre $(\text{a}\cos\alpha,\ \text{a}\sin\alpha)$ and radius a.
Answer
The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b)are centre and r is radius
From (A)
$(\text{x}-\text{a}\cos\alpha)^2+(\text{y}-\text{a}\sin\alpha)^2=\text{a}^2$
$\Rightarrow\text{x}^2-2\text{a}\cos\alpha\text{x}-2\text{a}\sin\alpha\ \text{y}=0$
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Question 81 Mark
If the abscissae and ordinates of two points $P$ and $Q$ are roots of the equations $x^2 + 2ax - b^2 = 0$ and $x^2 + 2px - q^2 = 0$ respectively, then write the equation of the circle with $PQ$ as diameter.
Answer
Let $\alpha,\ \beta$ abd $\gamma,\ \delta$ are the roots of the first and second given equation, so
$\alpha+\beta=-2\text{a}$ $\alpha\beta=-\text{b}^2$
$\gamma+\delta=-2\text{p}\ ,\ \gamma\delta=-\text{q}^2$
Coordinate of p and q are $(\alpha,\ \gamma)$ and $(\beta,\ \delta)$ respectively the equation of circle on PQ as diameter is
$(\text{x}-\alpha)(\text{x}-\beta)+(\text{y}-\gamma)(\text{y}-\delta)=0$
$\text{x}^2+\text{y}^2-(\alpha-\beta)\times-(\gamma+\delta)\text{y}+\alpha\beta+\gamma\beta=0$
$\text{x}^2+\text{y}^2+2\text{ax}+2\text{py}-\text{b}^2-\text{q}^2=0$
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Question 91 Mark
Find the equation of the circle with:
Centre (0, -1) and radius 1.
Answer
The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b) are centre and r is radius
From (A)
$(\text{x}-0)^2+(\text{y}+1)^2=1^2$
$\Rightarrow\text{x}^2+\text{y}^2+2\text{y}+1=1$
$\Rightarrow\text{x}^2+\text{y}^2+2\text{y}=0$
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Question 101 Mark
Write the equation of the unit circle concentric with $x^2 + y^2 - 8x + 4y - 8 = 0.$
Answer
$x^2 + y^2 - Bx + 4y - 8 = 0$
$x^2 - Bx + (4)^2 - (4)^2 + y^2 + 4y + (2)^2 - (2)^2 - 8 = 0$
$(x - 4)^2 + (y + 2)^2 - 16 - 4 - 8 = 0$
$(x - 4) + (y + 2) - 28 = 0$
$(x - 4)^2 + (y + 2)^2 = 28 .......... (1)$
Concentric unit circle to equation $(1)$ is
$(x - 4)^2 + (y + 2)^2 = 1$
$x^2 - Bx + y^2 + 4y + 19 = 0$
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Question 111 Mark
Write the coordinates of the centre of the circle passing through $(0, 0), (4, 0)$ and $(0, -6).$
Answer
Circle is passing through $(0, 0), (4, 0)$ and $(0, −6).$
Let equation of circle be,
$(x - h)^2 + (y - k)^2 = r^2 ...(1)$
Circle $(1)$ is passing through $A (0 , 0)$
$(0 - h)^2 + (0 - k)^2 = r^2$
$h^2 + k^2 = r^2 ...(2)$
Circle $(1)$ is passing through $B (4 , 0)$
$(4 - h)2 + (0 - k)2 = r2$
$(4 - h)^2 + k^2 = r^2 ...(3)$
Circle $(1)$ is passing through $B (0, -6)$
$(0 - h)^2 + (-6 - k)^2 = r^2$
$h^2 + (6 + k)^2 = r^2 ...(4)$
$[(2) - (3)],$
$h^2 - (4 - h)^2 = 0$
$(h - 4 + h) (h + 4 - h) = 0$
$(2h - 4) (4) = 0$
$h = 2$
$[(2) - (4)],$
$k^2 - (6 + k)^2 = 0$
$(k - 4 - k) (k + 6 + k) = 0$
$(-6) (2k + 6 = 0)$
$k = - 3$
put the value of h and k in equation $(1)$
$h^2 + k^2 = r^2$
$(2)^2 + (-3)^2 = r^2$
$r^2 = 13$
Centre of the circle $= (2, - 3)$
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Question 121 Mark
Find the equation of the circle with:
Centre (-2, 3) and radius 4.
Answer
The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}\ .....(\text{A})$
where(a, b)are centre and r is radius
$\therefore(\text{x}+2)^2+(\text{y}-3)^2=4^2$
$\Rightarrow(\text{x}+2)^2+(\text{y}-3)^2=16$
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Question 131 Mark
Find the centre and radius of the following circles:
$(\text{x}-1)^2+\text{y}^2=4$
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.
$(\text{x}-1)^2+\text{y}^2=4$
$(\text{x}-1)^2+\text{y}^2-4$
$\Rightarrow(\text{x}-1)^2+(\text{y}-0)^2-2^2$
comparing with (A) we get,
(1. 0) is the centre
2 ts the radius
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Question 141 Mark
Write the equation of the circle passing through $(3, 4)$ and touching y-axis at the origin.
Answer
Cirde is touching $y -$ axis at the origin.
Thus, centre of cirde is on $x-$ axis.
So Let centre of d rcle is $( h, o)$
Equation of circle is $(x - h)^2 + (y - o)^2 = r^2 ............ (1)$
It is passing through $(3, 4) (3 - h)^2 + (4)^2 = r^2 (3 - h)^2 + 16 = r^2 ............ (2)$
It is also passing through $(3, 4) (0 - h)^2 + (0 - o)^2 = r^2 h^2 = r^2$
Now, equation $(2)$ be com es, $(3 - h)^2 + 16 = h^2 9 - 6h + h^2 + 16 -6h + 25 = 0$
$\text{h}=\frac{25}{6}$Equation of circle is,
$(x - h)^2 + y^2 = h^2 x^2 - 2xh + y^2 = 0$
$\text{x}^2-2\text{x}\Big(\frac{25}{6}\Big)+\text{y}^2=0$
$6x^2 - 50x + 6y^2$
$= 0 3x^2 - 25x + 3y^{2}$
$ = 0 3(x^2 + y^2) - 25x$
$​​​​​​​= 0$
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Question 151 Mark
Find the equation of the circle with:
Centre (a, a) and radius $\sqrt{2}$ a.
Answer
The general equation of cirde is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b)are centre and r is radius
From (A)
$(\text{x}-\text{a})^2+(\text{y}-\text{a})^2=(\sqrt{2}\text{a})^2$
$\Rightarrow\text{x}^2-2\text{ax}+\text{a}^2+\text{y}^2-2\text{ay}+\text{a}^2=2\text{a}^2$
$\Rightarrow\text{x}^2+\text{y}^2-2\text{ax}-2\text{ay}=0$
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