5 Marks Questions

Question 15 Marks

Find the equations of the circle passing through the point $A(-4,2)$ and touching the lines $x+y=2$ and $x-y=2$.

Answer

The given lines are :
$\begin{array}{r}x+y-2=0\quad \ldots(i) \\ \text { and } x-y-2=0\quad \ldots(ii) \end{array}$

Let $C(h, k)$ be the centre of the circle which touches the line (i) and (ii) and passes through the point $A(-4,2)$, then
$\begin{array}{l} \text { distance of } C \text { from line (i) } =\text { distance of } C \text { from line (ii) } \\ =\text { radius of circle } \\ =C A .\end{array}$
$\begin{aligned} \Rightarrow \frac{|h+k-2|}{\sqrt{2}} & =\frac{|h-k-2|}{\sqrt{2}} \\ & =\sqrt{(h+4)^2+(k-2)^2}\end{aligned}$
$\Rightarrow |h+k-2|=|h-k-2|\quad \ldots(iii) $
and $|h+k-2|=\sqrt{2} \sqrt{(h+4)^2+(k-2)^2}\quad \ldots(iv) $
from eq. (iii), we get $h+k-2= \pm(h-k-2)$
$\Rightarrow h+k-2=h-k-2 \text { or } h+k-2=-h+k+2$
$\Rightarrow k=0 $ or $ h=2$
Case 1: When $k=0$, centre is $(h, 0)$ and
radius $=\sqrt{(h+4)^2+2^2}$ from eq. (iv), we get
$\Rightarrow |h-2|=\sqrt{2} \sqrt{(h+4)^2+(0-2)^2}$
$\Rightarrow h^2-4 h+4=2\left(h^2+8 h+16+4\right)$
$\Rightarrow h^2+20 h+36=0$
$\Rightarrow (h+2)(h+18)=0$
$\Rightarrow h=-2,-18$
When $h=-2$ then centre of circle is $(-2,0)$ and
$\text { radius }=\sqrt{(-2+4)^2+2^2}=2 \sqrt{2}$
The equation of the circle is $(x+2)^2+(y-0)^2=$ $(2 \sqrt{2})^2$
$\Rightarrow x^2+y^2+4 x-4=0$
When $h=-18$, then centre is $(-18,0)$ and
radius $=$ $\sqrt{(-18+4)^2+2^2}=10 \sqrt{2}$
The equation of the circle is $(x+18)^2+(y-0)^2=$ $(10 \sqrt{2})^2$
$\Rightarrow x^2+y^2+36 x+124=0$
Case II: When $h=2$, centre is $(2, k)$
From eq . (iii), we get
$|k|=\sqrt{2} \sqrt{(2+4)^2+(k-2)^2}$
$
\begin{array}{l}
\Rightarrow k^2=2\left(36+k^2-4 k+4\right) \\
\Rightarrow k^2-8 k+80=0
\end{array}
$
Its discriminant $=(-8)^2-4 \times 1 \times 80<0$, So this equation has no roots. So, there is no circle in this case.
Hence, the required equations of circles are :
$x^2+y^2+4 x-4=0 \text { and } x^2+y^2+36 x+124=0$

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Question 25 Marks

Show that the points $(7,5),(6,-2),(-1,-1)$ and $(0,6)$ are concyclic. Also, find the radius and the centre of the circle on which they lie.

Answer

First find the equation of the circle passes through the points $(7,5),(6,-2)$ and $(-1,-1)$.
Let the equation of circle be
$x^2+y^2+2 g x+2 f y+c=0\quad \ldots(i) $
As the points $(7,5),(6,-2)$ and $(-1,-1)$ lie on it, we get
$\begin{array}{lrl} 49+25+14 g+10 f+c =0 \\ \text { or, } 14 g+10 f+c+74 =0\quad \ldots(ii) \\ \text { and } 36+4+12 g-4 f+c =0 \\ \text { or, }12 g-4 f+c+40 =0\quad \ldots(iii) \\ \text { and }1+1-2 g-2 f+c =0 \\ \text { or, }2 g+2 f-c-2 =0\quad \ldots(iv) \end{array}$
and adding eqs. (ii) and (iv), we get
$\begin{aligned}16 g+12 f+72 =0 \\ \Rightarrow 4 g+3 f+18 =0\quad \ldots(v) \end{aligned}$
On adding eqs. (iii) and (iv), we get
$\begin{aligned} 14 g+2 f+38 =0 \\ \Rightarrow 7 g+ f+19 =0\quad \ldots(vi) \end{aligned}$
On solving eqs. (v) and (vi) simultaneously, we get
$g=-3, f=-2$
Substituting values of $g$ and $f$ in eq. (ii), we get
$c=-14(-3)-10(-2)-74=-12$
Now, substituting values of $g, f$ and $c$ in eq.(i), we get
$x^2+y^2-6 x-4 y-12=0\quad \ldots(vii) $
The fourth point $(0,6)$ will lie on eq. (vii), if
$0+36-0-24-12=0$
i.e., if $0=0$, which is true.
Hence, the given points are concyclic.
Also, eq. (vii) is the equation of the circle on which these points lie.
Its centre is $(3,2)$ and radius $=\sqrt{9+4-(-12)}=5$ unit

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Question 35 Marks

Find the equation of the circle passing through the vertices of a triangle whose sides are represented by the equations $x+y=2,3 x-4 y=6$ and $x-y=$ 0.

Answer

Let $A B, B C$ and $C A$ represent the three sides of the triangle given by $x+y=2,3 x-4 y=6$ and $x-y=$ 0 , respectively.
On solving $A B$ and $B C$, we get the coordinates of vertex $B(2,0)$.
On solving $B C$ and $C A$, we get $C(-6,-6)$.
and On solving $A B$ and $C A$, we get $A(1,1)$.

Let the equation of the circle be
$x^2+y^2+2 g x+2 f y+c=0\quad \ldots(i) $
Since, the circle (i) passes through $A(1,1), B(2,0)$ and $C(-6,-6)$, so, equation (i) implies
$\begin{aligned} 1^2+1^2+2 g \cdot 1+2 f \cdot 1+c & =0 \\ 2 g+2 f+c & =-2\quad \ldots(ii) \\ 2^2+0^2+2 g \cdot 2+2 f \cdot 0+c & =0 \\ 4 g+c+4 & =0\quad \ldots(iii) \end{aligned}$
and
$
\begin{array}{l}
(-6)^2+(-6)^2+2 g \cdot(-6)+2 f \cdot(-6)+c=0 \\
\Rightarrow12 g+12 f-c=72\quad \ldots(iv)
\end{array}
$
Solving equations (ii), (iii), (iv), we get
$g=2, f=3, c=-12$
Now, putting the value of $g, f, c$ in equation (i)
$x^2+y^2+4 x+6 y-12=0$
is the required equation of the circle.

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Question 45 Marks

$A(1,0)$ and $B(7,0)$ are two points on the axis of $X$. A point $P$ is taken in the first quadrant such that $P A B$ is an isosceles triangle and $P B=5$ units. Find the equation of the circle described on $P A$ as diameter.

Answer

Since, triangle is isosceles. (given)
The point $P$ lies on the perpendicular bisector of the base $A B$. So, the $X$-coordinate of $P$ is
$
\begin{aligned}
P(x, y) & =P\left(\frac{7+1}{2}, y\right) \\
& =P(4, y)
\end{aligned}
$

Now, $P A=5 \quad 1$
So, $(4-1)^2+y^2=25$
or, $y^2=25-9$
or, $y= \pm 41$
Since $P$ is in the first quadrant
$\therefore y=4$
So, coordinate of $P$ is $(4,4)$
Now, the equation of circle having PA as diameter using the diametric form as
$(x-4)(x-1)+(y-4)(y-0)=0$
$[\because$ Equation of circle having end points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ of diameter is $\left.\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0\right]$
$x^2+y^2-5 x-4 y+4=0$

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Question 55 Marks

The centre of a circle is in the first quadrant and the circle touches the $y$-axis at the point $(0,2)$ and passes through the point $(1,0)$. Find the equation of the circle.

Answer

Let $C$ be the centre of the circle. The circle touches the $Y$-axis at the point $B(0,2)$ and passes through the point $A(1,0)$
Since, the circle touches the $\gamma$-axis at the point $(0,2)$, So that the $y$-coordinate of the centre $C$ will be 2 .

Let the coordinates of the centre $C$ of the circle be $(a, 2)$.
Draw $C N$ perpendicular from $C$ upon the $X$-axis
Therefore, $O N=a$, so that
$A N=O N-O A=(a-1)$
Now, from the figure, it is clear that $C B=C A=$ radius of the circle $=a$
Therefore, from the right-angled triangle $A N C$, we have
$A C^2=A N^2+C N^2$
$\begin{array}{ll}\Rightarrow & a^2=(a-1)^2+(2)^2 \\ \Rightarrow & a^2=a^2-2 a+1+4 \\ \Rightarrow & 2 a=5 \\ \Rightarrow & a=\frac{5}{2}\end{array}$
Thus, the centre of the circle is $\left(\frac{5}{2}, 2\right)$ and its radius is $a=\frac{5}{2}$.
Hence, the required equation of the circle is
$\left(x-\frac{5}{2}\right)^2+(y-2)^2=\left(\frac{5}{2}\right)^2$
[ $\because$ equation of circle having centre $(h, k)$ and radius $a$ is $\left.(r-h)^2+(y-k)^2=a^2\right]$
$\begin{array}{lr}\Rightarrow\left(x^2-5 x+\frac{25}{4}\right)+\left(y^2-4 y+4\right)=\frac{25}{4} \\ \Rightarrow x^2+y^2-5 x-4 y+4=0\end{array}$

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Question 65 Marks

A circle passes through the points $(2,3)$ and $(3,-2)$. If the chord joining these points subtend a right angle at the centre, then find the equation of the circle.

Answer

Let the given points be $A(2,3)$ and $B(3,-2)$
Also, let the centre of the circle be $(h, k)$ and its radius be $a$ then the equation of the circle is

$\begin{aligned}(x-h)^2+(y-k)^2 =a^2 \\ \Rightarrow x^2+y^2-2 h x-2 k y+h^2+k^2 & =a^2\quad \ldots(i) \end{aligned}$
Since, the circle passes through the points $(2,3)$ and $(3,-2)$, we have
$\begin{array}{rlrl}13-4 h-6 k+h^2+k^2 & =a^2\quad \ldots(ii) \\ \text { and } 13-6 h+4 k+h^2+k^2 & =a^2\quad \ldots(iii) \end{array}$
Subtracting eq. (iii) from eq. (ii), we get
$2 h-10 k=0$
or, $h=5 k\quad \ldots(iv) $
Now, the gradients of the straight lines $A C$ and $B C$ are
$m_1=\frac{k-3}{h-2} \text { and } m_2=\frac{k+2}{h-3}$
Since, $\angle A C B=90^{\circ}$, we must have $m_1 m_2=-1$
Therefore,
$\frac{k-3}{h-2} \times \frac{k+2}{h-3}=-1$
$\begin{array}{lrl}\Rightarrow k^2-k-6 =-\left(h^2-5 h+6\right) \\ \Rightarrow h^2+k^2-k-5 h =0 \\ \Rightarrow 25 k^2+k^2-k-25 k =0\quad \ldots(v) \end{array}$
[using eq. (iv)]
$\begin{aligned} \Rightarrow & 26 k^2-26 k =0 \\ \Rightarrow & k(k-1) =0 \\ \Rightarrow & k =0 \text { or } k=1\end{aligned}$
When $k=0$, then eq. (iv) gives $h=0$ and then from eq. (ii), $a^2=13$
$\therefore$ From eq. (i), equation of the circle is
$x^2+y^2=13$
Again, when $k=1$, then from eq. (iv), $h=5$ and then from eq (ii), we get
$13-20-6+25+1=a^2$
or $a^2=13$
Hence, in this case, the required equation of the circle is
$x^2+y^2-10 x-2 y+13=0$

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Applied Maths 5 Marks Questions

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