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The Circle · Gujarat Board (GSEB) STD 11 Science (GSEB - 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=m x \text { does not intersect the circle }(x+10)^2+(y+10)^2=180 \text { Now, }(x+10)^2+(m x+10)^2=180 x^2+100+20 x+ $
$m^2 x^2+100+20 m x=180 x^2+m^2 x^2+20 m x+20 x+20=0 x^2\left(1+m^2\right)+20 x(m+1)+20=0 \text { Since } y=m x \text { does } $
$\text { not intersect so } b^2-4 a c<0\left[20(m+1)^2\right]-4\left(1+m^2\right)(20)<0400(m+1)^2-80\left(1+m^2\right)<05 m^2+5+10 m-1-m^2< $
$04 m^2+10 m+4<04 m^2+8 m+2 m+4<04 m(m+2)+2(m+2)<0(m+2)(4 m+2)<0$
$\Rightarrow m+2<0 \text { and } 4 m $
$+2>0$
$\Rightarrow m<-2 \text { and } m>-2 \text { Its not possible }$
$\Rightarrow m+2>0 \text { and } 4 m+2<0$
$\Rightarrow m>-2 \text { and } m<-\frac{1}{2} $
$\Rightarrow-2<m<-\frac{1}{2}$
$\Rightarrow m \in\left(-2, \frac{1}{2}\right)$
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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
$x^2+y^2-B x+4 y-8=0 x^2-B x+(4)^2-(4)^2+y^2+4 y+(2)^2-(2)^2-8=0(x-4)^2+(y+2)^2-16-4-8=0(x-4)+(y$
$+2)-28=0(x-4)^2+(y+2)^2=28$ $\qquad$ (1) Concentric unit circle to equation (1) is $(x-4)^2+(y+2)^2=1 x^2-B x+$ $y^2+4 y+19=0$
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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, $(x-h)^2+(y-k)^2=r^2 \ldots(1)$ Circle (1) is passing through $A(0,0)(0-h)^2+(0-k)^2=r^2 h^2+k^2=r^2 \ldots(2)$ Circle (1) is passing through $B(4,0)(4-h) 2+(0-$ k) $2=r 2(4-h)^2+k^2=r^2 \ldots(3)$ Circle (1) is passing through $B(0,-6)(0-h)^2+(-6-k)^2=r^2 h^2+(6+k)^2=r^2 \ldots(4)[(2)-$ (3)], $h^2-(4-h)^2=0(h-4+h)(h+4-h)=0(2 h-4)(4)=0 h=2[(2)-(4)], k^2-(6+k)^2=0(k-4-k)(k+6+k)=$
$0(-6)(2 k+6=0) k=-3$ 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 rele 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, $(3-h)^2+16=h^2 9-6 h+h^2+16-6 \mathrm{~h}+25=0 \mathrm{~h}=\frac{25}{6}$ Equation of circle is, $(x-h)^2+y^2=h^2 x^2-2 x h+y^2=0 x^2-2 x\left(\frac{25}{6}\right)+y^2=06 x^2-50 x+6 y^2=03 x^2-25 x+3 y^2=03\left(x^2+y^2\right)-25 x=0$
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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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(Each question 1 marks) - MATHS STD 11 Science Questions - Vidyadip