Circle | Part-5 | S N Dey Examhoop

In the previous article , we discussed few Short Answer Type Questions. In this article, we will discuss 9 more Short Answer type Questions from Chhaya Mathematics , Class 11 (S N De book ).

Short Answer Type Questions of Circle, S N Dey Mathematics, Class 11

21. A circle touches $~y-$ axis at $~(0,5)~$ and whose centre lies on the line $~2x+y=13; ~$ find the equation of the circle.

Solution.

Let the radius of the circle be $~r~$ unit. The circle touches the $~y-$ axis at the point $~(0,5).$ Now, since $~ AB \perp OY~$ where $~A(0,5)~$ and the centre of the circle is $~B(r,5).$

Since the centre of the circle lies on $~2x+y=13,~$

so $~2r+5=13 \Rightarrow r=\frac{13-5}{2}=\frac{8}{2}=4.$

$\therefore~$ the centre of the circle is $~(4,5)~$ and the radius is $~4~$ unit.

$\therefore~$ the equation of the circle is

$(x-4)^2+(y-5)^2=4^2 \\ \text{or,}~~ x^2-2 \times x \times 4+4^2+y^2-2 \times y \times 5+5^2=4^2 \\ \text{or,}~~ x^2+y^2-8x-10y+25=0.$

22. Find the equation of a circle which passes through the point $~(4,2)~$ and touches both the co-ordinate axes. How many of such circles are possible ?

Solution.

Let the radius of the circle be $~r~$ unit and the circle touches $~x~$ axis and $~y~$ axis at $~A~$ and $~B~$ respectively.

$\therefore~ AC \perp OA,~~ BC \perp OB. \ \therefore ~ AC=BC=r$

$\therefore~$ the co-ordinates of $~C~$ is $~(r, r).$

So, the equation of the circle can be written as

$~(x-r)^2+(y-r)^2=r^2 \rightarrow(1)$ which passes through $~(4,2).$

$\therefore~(4-r)^2+(2-r)^2=r^2 \\ \text{or,}~~ 16-8r+r^2+4-4r+r^2=r^2 \\ \text{or,}~~ r^2-12r+20=0 \\ \text{or,}~~ r^2-10r-2r+20=0 \\ \text{or,}~~r(r-10)-2(r-10)=0 \\ \text{or,}~~ (r-10)(r-2)=0 \\ \text{or,}~~r=2,10 \rightarrow(2).$

Finally, by $~(1)~$ and $~(2)~$ , the equation of circle can be written as

$(x-2)^2+(y-2)^2=2^2=4~$ or, $~(x-10)^2+(y-10)^2=10^2=100.$

So, from the aforementioned discussions we can conclude that two such circles are possible.

23. Two circles of radii $~2~$ and $~10~$ units respectively pass through $~(2,4)~$ and touch both the $~x~$ and $~y~$ axes. Find the equations of the two circles . Also find the other common point of intersection.

Solution.

Since the circles passes through the point $~(2,4)~$ which lies in the first quadrant and since the given circles touch both the $~x~$ and $~y~$ axes, so the circles lie in the first quadrant.

Suppose that the circle with radius $~2~$ unit touches $~x~$ axis at the point $~A~$ and touches $~y~$ axis at $~B~$ .

$\therefore~ CB \perp OB,~~ AC \perp OA.$

Now from the following figure, we can notice that the co-ordinates of $~C~$ is $~(2,2).$

So, the equation of the circle

$~(x-2)^2+(y-2)^2=2^2 \\ \text{or,}~~ x^2-2x+4+y^2-4y+4=4 \\ \therefore x^2+y^2-4x-4y+4=0 \rightarrow(1)$

Similarly, we can determine that the equation of the circle with radius $~10~$ unit is given by

$~(x-10)^2+(y-10)^2=10^2 \\ \text{or,}~~ x^2-20x+100+y^2-20y+100=100 \\ \therefore x^2+y^2-20x-20y+100=0\rightarrow(2)$

Now, we need to determine the other common point intersection.

Subtracting $~(2)~$ from $~(1)~$ we get,

$16x+16y-96=0 \\ \text{or,}~~ 2x^2-12x+16=0 \\ \text{or,}~~ x^2-6x+8=0 \\ \text{or,}~~ x^2-4x-2x+8=0 \\ \text{or,}~~ x(x-4)-2(x-4)=0 \\ \text{or,}~~ (x-4)(x-2)=0 \\ \text{or,}~~x=2,4 \rightarrow(4)$

From $~(3)~$ and $~(4)~$ we get that for $~x=2,4~$ the corresponding values of $~y=4,2.$

So, common points of intersection of two circles are given by $~(2,4)~$ and $~(4,2).$

Hence, the other common point of intersection is $~(4,2).$

24. Prove that the $~(-1,-2)~$ lies on the circle $~x^2+y^2-x-y-8=0.~$ Find the co-ordinates of the other extremity of the diameter through $~(-1,-2).$

Solution.

Putting $~x=-1,~y=-2~$ in the equation of the circle $~x^2+y^2-x-y-8=0,~$ we get $~(-1)^2+(-2)^2-(-1)-(-2)-8=0~$ and so the point $~(-1,-2)~$ lies on the given circle.

Comparing $~x^2+y^2-x-y-8=0,~$ with the general form of the circle $~x^2+y^2+2gx+2fy+c=0~$ we get,

$~2g=-1 \Rightarrow g=-\frac 12,~~ 2f=-1\Rightarrow f=-1/2.$

So, the centre of the given circle $~(1/2,1/2).$

Let the other extremity of the circle be $~(\alpha,\beta).$

Since the the point $~(-1,-2)~$ lies on the circle, so the centre of the circle will be the midpoint of the straight line joining the points $~(-1,-2)~$ and $~(\alpha,\beta),~$ which is $~\left(\frac{\alpha-1}{2},\frac{\beta-2}{2}\right).$

$\text{So,}~\frac{\alpha-1}{2}=\frac 12 \Rightarrow \alpha=2 ;~~\frac{\beta-2}{2}=\frac 12 \Rightarrow \beta=3.$

Hence, the co-ordinates of the other extremity of the diameter through $~(-1,-2)~$ is $~(2,3).$

25. Show that for all values of $~p~$ , the circle $~x^2+y^2-x(3p+4)-y(p-2)+10p=0~$ passes the point $~(3,1).~$ If $~p~$ varies , find the locus of the centre of the above circle.

Solution.

Since the given circle $~x^2+y^2-x(3p+4)-y(p-2)+10p=0\rightarrow(1)~$ passes the point $~(3,1),~$ putting $~x=3,~y=1~$ in $~(1),~$ we get

$~3^2+1^2-3(3p+4)-1(p-2)+10p\\=10-9p-12-p+2+10p\\=0.$

So, $~x=3,y=1~$ satisfy the given equation of the circle. So, the circle passes through the point $~(3,1)~$ for any values of $~p.$

Comparing $~x^2+y^2-x(3p+4)-y(p-2)+10p=0~$ with the general form of the circle $~x^2+y^2+2gx+2fy+c=0~$ we get,

$~2g=-(3p+4) \Rightarrow g=-\frac{3p+4}{2},~~ 2f=-(p-2)\Rightarrow f=-\frac{p-2}{2}$

So, the centre of the circle $~(-g,-f)=\left(\frac{3p+4}{2},\frac{p-2}{2}\right)$

If $~(h,k)~$ is the centre of the circle, then

$h=\frac{3p+4}{2}~$ and $k=\frac{p-2}{2} \rightarrow(2)$

From $~(2),~$ we get $~p=\frac 13(2h-4),~~p=2k+2.$

$\therefore \frac 13(2h-4)=2k+2 \\ \text{or,}~~ 2h-6k-10=0 \\ \text{or,}~~ 2(h-3k-5)=0 \\ \text{or,}~~ h-3k-5=0 \rightarrow(3)$

Finally, by $~(3),~$ we can conclude that the locus of the centre of the circle is $~ x-3y-5=0.$

26. Find the co-ordinates of the points equidistant from the axes and lying on the circle $~x^2+y^2-6x-2y+6=0.$

Solution.

The given equation of circle can be rewritten as :

$x^2+y^2-6x-2y+6=0 \\ \text{or,}~~ (x^2-6x+9)+(x^2-2y+1)=4 \\ \text{or,}~~ (x-3)^2+(y-1)^2=2^2$

Let the co-ordinates of the points equidistant from the axes be $~(a,a),~~(a,-a).$

Since the point $~(a,a)~$ lies on the given circle,

$a^2+a^2-6a-2a+6=0 \\ \text{or,}~~2a^2-8a+6=0 \\ \text{or,}~~ 2(a^2-4a+3)=0 \\ \text{or,}~~ a^2-4a+3=0 \\ \text{or,}~~ a^2-a-3a+3=0 \\ \text{or,}~~a(a-1)-3(a-1)=0 \\ \text{or,}~~ (a-1)(a-3)=0 \\ \text{or,}~~ a=3,1.$

The co-ordinates of the points equidistant from the axes are $~(3,3),~(1,1).$

Again, since the point $~(a,-a)~$ lies on the given circle,

$a^2+a^2-6a+2a+6=0 \\ \text{or,}~~ 2a^2-4a+6=0 \\ \text{or,}~~ 2(a^2-2a+3)=0 \\ \text{or,}~~ a^2-2a+3=0 \\ \text{or,}~~ a=\frac{-(-2) \pm \sqrt{(-2)^2-4 \times 1 \times 3}}{2} \\ \text{or,}~~ a=1 \pm \sqrt{-2} \rightarrow(1)$

So, by $~(1)~$ we get the imaginary values of $~a~$ and so we can discard those values.

27. Find the equation to the common chord of the two circles $~x^2+y^2-4x+6y-36=0~$ and $~x^2+y^2-5x+8y-43=0~$ and also find its length.

Solution.

The equation to the common chord of the two circles $~x^2+y^2-4x+6y-36=0~$ and $~x^2+y^2-5x+8y-43=0~$ is

$(x^2+y^2-4x+6y-36)-(x^2+y^2-5x+8y-43)=0 \\ \text{or,}~~ x-2y+7=0\rightarrow(1)$

From $~(1)~$ we get, $~x=2y-7 \rightarrow(7)$

By $~(1)~$ and $~(2)~$ we get by replacing the value of $~x,$

$(2y-7)^2+y^2-4(2y-7)+6y-36=0 \\ \text{or,}~~ 4y^2-28y+49+y^2-8y+28+6y-36=0 \\ \text{or,}~~ 5y^2-30y+41=0 \rightarrow(3)$

Let the roots of the equation in $~(3)~$ be $~y_1,y_2.$

$~\therefore y_1+y_2=\frac{30}{5}=6~~$ and $~y_1y_2=\frac{41}{5}$

$(y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2 \\ \text{or,}~~ (y_1-y_2)^2=36- \frac{4 \times 41}{5} \\ \text{or,}~~ (y_1-y_2)^2=\frac{16}{5}\rightarrow(4)$

Let the chord intersects the circles at the points $~(x_1,y_1)~$ and $~(x_2,y_2)~$ .

$~ \therefore x_1=2y_1-7~$ and $x_2=2y_2-7.$

So, the length of the chord is

28. Find the equation of the common chord of the two circles $~x^2+y^2-4x-10y-7=0~$ and $~2x^2+2y^2-5x+3y+2=0.~$ Show that this chord is perpendicular to the line joining the centres of two circles.

Solution.

The equation of two given circles are $~x^2+y^2-4x-10y-7=0 \rightarrow(1)$ and $~2x^2+2y^2-5x+3y+2=0 \Rightarrow x^2+y^2-(5/2)x+(3/2)y+1=0\rightarrow(2)$

Now, subtracting $~(1)~$ from $~(2),~$ we get

$~(3/2)x+\frac{23}{2}y+8=0 \\ \text{or,}~~ 3x+23y+16=0 \rightarrow(3)$

So, equation in $~(3)~$ represents the equation of the common chord of two circles.

The co-ordinates of centre of circle $~(1)~$ is $~(2,5)~$ and the co-ordinates of centre of circle $~(2)~$ is $~\left(\frac 54,-\frac 34\right).$

Now, the slope of the straight line joining two centres of the circles is $~\frac{5+\frac 34}{2-\frac 54}=\frac{23}{3}.$

Again , the slope of the straight line $~(3)~$ is $~-\frac{3}{23}.$

So, $~~\left(-\frac{3}{23}\right)\times \frac{23}{3}=-1\rightarrow(4).$

Finally, by $~(4),~$ we can conclude that this chord is perpendicular to the line joining the centres of two circles.

29. Find the equation of the circle which passes through the origin and the points of intersection of the circles $~~x^2+y^2-4x-8y+16=0~$ and $~x^2+y^2+6x-4y-3=0.$

Solution.

The equation of the circle through the points of intersection of the circles $~~x^2+y^2-4x-8y+16=0~$ and $~x^2+y^2+6x-4y-3=0~$ is

$x^2+y^2-4x-8y+16 +k(x^2+y^2+6x-4y-3)=0 \\ \text{or,}~~ (1+k)x^2+(1+k)y^2+(6k-4)x-(8+4k)y+16-3k=0 \rightarrow(1)$

Since the circle $~(1)~$ passes through the origin, putting $~x=0,~y=0~$ in $~(1)~$ we get,

$~6-3k=0 \Rightarrow k=\frac{16}{3}.$

$\therefore~$ the equation of the circle is

$\left(1+\frac{16}{3}\right)x^2+\left(1+\frac{16}{3}\right)y^2+\left(\frac{6 \times 16}{3}-4\right)x-\left(8+\frac{4 \times 16}{3}\right)y=0 \\ \text{or,}~~ \frac{19}{3}x^2+\frac{19}{3}y^2+\frac{84}{3}x-\frac{88}{3}y=0 \\ \text{or,}~~ 19(x^2+y^2)+84x-88y=0.$

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Short Answer Type Questions of Circle, S N Dey Mathematics, Class 11

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