Circle | Part-8 | S N Dey Examhoop

In the previous article , we discussed about 10 Long Answer Type Questions. In this article, we will discuss few more Long Answer type Questions from Chhaya Mathematics , Class 11 (S N De book ).

Circle Related Long Answer Type Questions (11-20) | S N Dey Mathematics

11. Show that the circle $~x^2+y^2+6(x-y)+9=0~$ touches the co-ordinate axes. Also find the equation of the circle which passes through the common points of intersection of the above circle and the straight line $~x-y+4=0~$ and which also passes through the origin.

Solution.

The equation of the given circle can be written as $~(x+3)^2+(y-3)^2=3^2 \rightarrow(1)$

From $~(1),~$ we get the centre of the circle $~(1)~$ is $~(-3,3)~$ and radius $~3~$ unit.

Now, if $~x=0~$ , we get from the given equation of the circle,

$~y^2-6y+9=0 \Rightarrow (y-3)^2=0 \Rightarrow y=3,3.$

As we have just one point common between $~y-$ axis and circle, it means it touches
$~y-$ axis at $~(0,3).$

Similarly, if $~y=0~$ , we get from the given equation of the circle,

$~x^2+6x+9=0 \Rightarrow (x+3)^2=0 \Rightarrow y=-3,-3.$

As we have just one point common between $~x-$ axis and circle, it means it touches
$~x-$ axis at $~(-3,0).$

Hence, the given circle touches the two axes.

Now, the equation of the circle passing through the intersection of the given circle and the straight line $~x-y+4=0~$ is

$~(x^2+y^2+6x-6y+9)+k(x-y+4)=0 \rightarrow(2)$

Since the circle $~(2)~$ passes through the point $~(0,0)~$ , we get from $~(2),$

$9+k \times 4=0 \Rightarrow k=-\frac 94.$

Hence, the equation of the circle is

$x^2+y^2+6x-6y+9-\frac 94(x-y+4)=0 \\ \text{or,}~~ 4(x^2+y^2)+24x-24y+36-9x+9y-36=0 \\ \text{or,}~~4(x^2+y^2)+15(x-y)=0$

12. A circle through the common points of the circles $~x^2+y^2-2x-4y+1=0~$ and $~x^2+y^2+2x-6y+1=0~$ has its centre on the line $~4y-7x-19=0.~$ Find the centre and radius of the circle.

Solution.

Any circle through the common points of the circles $~x^2+y^2-2x-4y+1=0~$ and $~x^2+y^2+2x-6y+1=0~$ can be written as

$(x^2+y^2-2x-4y+1)+k(x^2+y^2+2x-6y+1)=0 \\ \text{or,}~~(1+k)x^2+(1+k)y^2-2(1+k)x-2y(2+3k)+(1+k)=0 \\ \text{or,}~~ x^2+y^2-2x-2y \left(\frac{2+3k}{1+k}\right)+1=0 \rightarrow(1)$

Comparing $~(1)~$ with $~x^2+y^2+2gx+2fy+c=0~$ , we get

$2g=-2 \Rightarrow -g=1, \\~~ 2f=-2\left(\frac{2+3k}{1+k}\right) \Rightarrow -f=\frac{2+3k}{1+k},~~ c=1.$

So, centre of the circle $~(1)~$ is $~(-g,-f)=\left(1,\frac{2+3k}{1+k} \right)$

Since the centre of the circle $~(1)~$ lies on the line $~4y-7x-19=0~$ ,

$4\left(\frac{2+3k}{1+k} \right)-7 \times 1-19=0 \\ \text{or,}~~4(3k+2)-26(k+1)=0 \\ \text{or,}~~ 12k+8-26k-26=0 \\ \text{or,}~~-14k=18 \\ \text{or,}~~ k=-\frac{18}{14}=-\frac 97.$

Hence, the centre of the circle $~(1)~$

$\left(1,\frac{2+3k}{1+k} \right)\\~~=\left(1,\frac{2+3(-9/7)}{1-9/7}\right)\\~~=\left(1,\frac{-13/7}{-2/7}\right)\\~~=\left(1,\frac{13}{2}\right)$

Also, the radius of the circle is

$=\sqrt{g^2+f^2-c}=\sqrt{1^2+(13/2)^2-1}=\frac{13}{2}~~\text{unit}.$

13. The circle $~x^2+y^2+2x-4y-11=0~$ and the line $~x-y+1=0~$ intersect at $~A~$ and $~B~$ . Find the equation of the circle on $~AB~$ as diameter.

Solution.

The equation of the given straight line $~x-y+1=0 \rightarrow(1).$

The equation of the given circle is $~x^2+y^2+2x-4y-11=0\rightarrow(2)$

From $~(1)~$ we get, $~x=y-1 \rightarrow(3)$

From $~(2)~$ and $~(3)~$ we get,

$(y-1)^2+y^2+2(y-1)-4y-11=0 \\ \text{or,}~~y^2-2y+1+2y-2-4y-11=0 \\ \text{or,}~~ 2y^2-4y-12=0\\ \text{or,}~~ 2(y^2-2y-6)=0 \\ \text{or,}~~ y^2-2y-6=0 \\ \text{or,}~~ (y-1)^2-7=0 \\ \text{or,}~~ y-1=\pm\sqrt{7} \\ \text{or,}~~ y=\pm\sqrt{7}+1 \rightarrow(4)$

Now, by $~(3),~(4)~$ we get for $~y=\sqrt{7}+1,~~x=\sqrt{7}+1-1=\sqrt{7}.$

Similarly, for $~y=-\sqrt{7}+1,~~x=-\sqrt{7}+1-1=-\sqrt{7}.$

$\therefore~ A \equiv (\sqrt{7}, \sqrt{7}+1), ~~ B\equiv (-\sqrt{7},-\sqrt{7}+1).$

So, the equation of the circle on $~AB~$ as diameter is given by

$(x-\sqrt{7})(x+\sqrt{7})+[y-(\sqrt{7}+1)][y-(-\sqrt{7}+1)]=0 \\ \text{or,}~~ x^2-(\sqrt{7})^2+[(y-1)-\sqrt{7}][(y-1)+\sqrt{7}]=0 \\ \text{or,}~~ x^2-7+(y-1)^2-(\sqrt{7})^2=0 \\ \text{or,}~~ x^2-7+y^2-2y+1-7=0 \\ \therefore x^2+y^2-2y-13=0.$

14. Find the equation to the circle described on the common chord of the circles $~x^2+y^2-4x-2y-31=0~$ and $~2x^2+2y^2-6x+8y-35=0~$ as a diameter.

Solution.

Two given equations of circles are

$x^2+y^2-4x-2y-31=0~\rightarrow(1), \\ ~~ 2x^2+2y^2-6x+8y-35=0 \\ \text{or,}~~ x^2+y^2-3x+4y-35/2=0 \rightarrow(2).$

So, by $~(2)-(1),~$ we get the equation of common chord which is $~ x+6y+\frac{27}{2}=0 \rightarrow(3)$

Equation of any circle through the intersection of two given circles are

$x^2+y^2-4x-2y-31+k(x^2+y^2-3x+4y-35/2)=0 \rightarrow(4) \\ \text{or,}~~ x^2+y^2-x \cdot \frac{4+3k}{1+k}+2y \cdot \frac{2k-1}{1+k}-\frac{31+\frac{25}{2}k}{1+k}=0\rightarrow(5)$

From $~(5),~$ we get the centre of circle $\left(\frac{4+3k}{2(k+1)},\frac{-2k+1}{k+1}\right)~$ which lies on the common chord $~(3).$

$\text{So,}~~ \frac{4+3k}{2(k+1)} + 6 \times \frac{-2k+1}{k+1}+\frac{27}{2}=0 \\ \text{or,}~~ 4+3k+12(-2k+1)+27(k+1)=0 \\ \text{or,}~~ k=-\frac{43}{6}$

Now, putting the value of $~k~$ in $~(4),~$ we get

$x^2+y^2-4x-2y-31-\frac{43}{6}(x^2+y^2-3x+4y-35/2)=0 \\ \text{or,}~~ 2(x^2+y^2-4x-2y-31)-\frac{43}{6}(2x^2+2y^2-6x+8y-35)=0 \\ \text{or,}~~ 74(x^2+y^2)-210x+368y-1133=0.$

15. Find the equation to the locus of the mid-points of chords drawn through the point $~(0,4)~$ on the circle $~x^2+y^2=16.$

Solution.

Let $~AP~$ be the chord passing through the point $~A(0,4).$

Any point $~(P)~$ on the circle $~x^2+y^2=16=4^2~$ can be written as $~P(4\cos\theta,4\sin\theta).$

If $~(h,k)~$ is the co-ordinates of mid-point of $~AP~$ , then

$h=\frac 12(4\cos\theta+0),~~k=\frac 12(4+4\sin\theta). \\ \therefore~~ \cos\theta=\frac h2,~~\sin\theta=\frac 12(k-2)\rightarrow(1).$

Then from $~(1)~$ we get,

$\sin^2\theta+\cos^2\theta=1 \\ \text{or,}~~[\frac 12(k-2)]^2+(h/2)^2=1 \\ \text{or,}~~ h^2+(k-2)^2=4 \\ \text{or,}~~ h^2+k^2-4k+4=4 \\ \text{or,}~~ h^2+k^2=4k \rightarrow(2)$

Hence, by $~(2),~$ we can say that the equation to the locus of the mid-points of chords is $~x^2+y^2=4y.$

16. A circle passes through the origin $~O~$ and intersects the co-ordinate axes at $~A~$ and $~B~$ . If the length of the diameter of the circle be $~3c~$ unit, then find the locus of the centroid of the triangle $~OAB.$

Solution.

Any circle passing through the origin can be written as $~x^2+y^2+2gx+2fy=0 \rightarrow(1).$

Let $~A \equiv (a,0);~~B \equiv (0,b).$

Since the circle $~(1)~$ passes through the points $~A,B~$ so

$~a^2+2ga=0 \Rightarrow g=-a/2, ~\text{and}~b^2+2fb=0 \Rightarrow f=-b/2.$

$\therefore~ \sqrt{g^2+f^2}=\frac{3c}{2} \\ \text{or,}~\sqrt{(-a/2)^2+(-b/2)^2}=3c/2 \\ \text{or,}~~ \frac 12 \sqrt{a^2+b^2}=\frac 12 \times 3c \\ \text{or,}~~ \sqrt{a^2+b^2}=3c \rightarrow(2)$

If $~(h,k)~$ be the centroid of $~\Delta OAB~$ , then

$~h=a/3,~~k=b/3 \rightarrow(3)$

So, by $~(2),~(3)~$ we get,

$a^2+b^2=9c^2 \\ \text{or,}~~ (3h)^2+(3k)^2=9c^2 \\ \text{or,}~~ h^2+k^2=c^2.$

$\therefore~$ the locus of the centroid of the triangle $~OAB~$ is $~x^2+y^2=c^2.$

17. Find the equation of a circle circumscribing the triangle whose sides are $~x=0,~y=0~$ and $~lx+my=1.$ If $~l,m~$ vary so that $~l^2+m^2=4l^2m^2~$ , find the locus of the centre of the circle.

Solution.

18.Find the area of the equilateral triangle inscribed in the circle $~x^2+y^2+2gx+2fy+c=0.$

Solution.

The equation of the given circle is $~x^2+y^2+2gx+2fy+c=0 \rightarrow(1).$

The centre of the circle $~(1)~$ is $~(-g,-f)~$ and radius $~(r)=\sqrt{g^2+f^2-c}.$

Now, from the figure , we note that in $~\Delta ADC,$

$\cos 30^{\circ}=\frac{a/2}{r} \\ \text{or,}~~ \frac{\sqrt{3}}{2}=\frac{a}{2r} \\ \text{or,}~~ a=\sqrt{3}r.$

So, the area of the equilateral triangle inscribed in the circle is

$=\frac{\sqrt{3}}{4}a^2\\~~~=\frac{\sqrt{3}}{4} \cdot (\sqrt{3}r)^2 \\~~~=\frac{\sqrt{3}}{4} \cdot 3(g^2+f^2-c)\\~~~=\frac{3\sqrt{3}}{4}(g^2+f^2-c).$

19. Find the area of the equilateral triangle inscribed in the circle $~x^2+y^2-4x+6y-3=0.$

Solution.

Comparing the given circle with the standard form of the circle $~x^2+y^2+2gx+2fy+c=0~$ we get,

$2g=-4 \Rightarrow g=-\frac 42=-2, ~~ 2f=6\Rightarrow f=3,~~ c=-3.$

So, the radius $~(r)=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+3^2+3}=4~~\text{unit.}$

Now, from the figure ( $\Delta ABC$ ) we notice that

$OB=r, ~~ \angle{OBM}=30^{\circ},~~ \angle{BOM}=60^{\circ}.$

$\text{Now,}~~ BM=OB\sin 60^{\circ}=\frac{\sqrt{3}}{2}r \\ \therefore~ BM=\frac{\sqrt{3}}{2} \times 4=2\sqrt{3}~\text{unit}.$

So, the length $~(a)~$ of the side of $~\Delta ABC~$ is $BC=2BM=2 \times 2\sqrt{3}=4\sqrt{3}~~\text{unit}.$

So, the area of the equilateral triangle inscribed in the circle

$=\frac{\sqrt{3}}{4} a^2\\=\frac{\sqrt{3}}{4} \times (4\sqrt{3})^2\\=\frac{\sqrt{3}}{4} \times 48\\=12\sqrt{3}~~\text{(sq. unit)}$

20. Prove analytically that the straight line joining the middle point of a chord of a circle of a circle with the centre is perpendicular to the chord.

Solution.

To prove : $~OC=AB.~~$ Given that $~AC=CB.$

From $~\Delta OAC~$ and $~\Delta OBC~$ , we get

$OA=OB,~OC=OC,~AC=CB ~~\text{(Given)}$

$\therefore~\Delta OAC \cong \Delta OBC~~(S-S-S)$

$\therefore~ \angle{OCA}=\angle{OCB}. \\~~\\~~ \text{Now,}~\angle{OCA}+\angle{OCB}=180^{\circ} \\~ \text{or,}~~2\angle{OCA}=180^{\circ} \\ \text{or,}~~\angle{OCA}=180^{\circ}/2=90^{\circ} \\ \therefore~\angle{OCA}=\angle{OCB}=90^{\circ}.$

Hence, $~OC \perp AB.$

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Circle Related Long Answer Type Questions (11-20) | S N Dey Mathematics

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