Answer the following questions in short.

Question 11 Mark

Show that $2 x+y+6=0$ is a tangent to $x^2+y^2+2 x-2 y-3=0$. Find its point of contact.

Answer

Given equation of circle is

$x^2+y^2+2 x-2 y-3=0 \ldots(i)$

Given equation of line is

2x + y + 6 = 0 y = -6 – 2x ……(ii)

Substituting y = -6 – 2x in (i), we get

$\begin{aligned} & x+(-6-2 x)^2+2 x-2(-6-2 x)-3=0 \\ & \Rightarrow x^2+36+24 x+4 x^2+2 x+12+4 x-3=0 \\ & \Rightarrow 5 x^2+30 x+45=0 \\ & \Rightarrow x^2+6 x+9=0 \\ & \Rightarrow(x+3)^2=0 \\ & \Rightarrow x=-3\end{aligned}$

Since, the roots are equal.

$\therefore 2 x+y+6=0$ is a tangent to $x^2+y^2+2 x-2 y-3=0$

Substituting x = -3 in (ii), we get

y = -6 – 2(-3) = -6 + 6 = 0

Point of contact = (-3, 0)

View full question & answer→

Question 21 Mark

Find the equation of a circle with centre at origin and radius 3.

Answer

Standard equation of a circle is
$
\therefore \begin{aligned}
x^2+y^2 & =r^2 \quad \text { here } r=3 \\
x^2+y^2 & =3^2 \\
x^2+y^2 & =9 \quad \text { is the equation of circle }
\end{aligned}
$

View full question & answer→

Question 31 Mark

Find the centre and radius of the following circles:
$\left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{3}\right)^2=\frac{1}{36}$

Answer

Given the equation of the circle is
$\left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{3}\right)^2=\frac{1}{36}$
$\left(x-\frac{1}{2}\right)^2+\left(y-\left(-\frac{1}{3}\right)\right)^2=\left(\frac{1}{6}\right)^2$
Comparing this equation with $(x-h)^2+(y-k)^2=r^2$, we get
$h=\frac{1}{2}, k=\frac{-1}{3}$ and $r=\frac{1}{6}$
Centre of the circle $=(\mathrm{h}, \mathrm{k})=\left(\frac{1}{2}, \frac{-1}{3}\right)$ and radius of the circle $=\frac{1}{6}$

View full question & answer→

Question 41 Mark

Find the centre and radius of the following circles:
$(x-5)^2+(y-3)^2=20$

Answer

Given equation of the circle is
$(x-5)^2+(y-3)^2=20$
$\Rightarrow(x-5)^2+(y-3)^2=(\sqrt{ } 20)^2$
Comparing this equation with $(x-h)^2+(y-k)^2=r^2$, we get
$h = 5, k = 3$ and $r = \sqrt{20} = 2\sqrt{5}$
Centre of the circle $= (h, k) = (5, 3)$
and radius of the circle $= 2\sqrt5.$

View full question & answer→

Question 51 Mark

Find the centre and radius of the following circles:
$x^2+y^2=25$

Answer

Given equation of the circle is
$x^2+y^2=25$
$\Rightarrow x^2+y^2=(5)^2$
Comparing this equation with $x^2+y^2=r^2$, we get $r=5$
Centre of the circle is $(0, 0)$ and radius of the circle is $5.$

View full question & answer→

Maths Answer the following questions in short.

Take a timed test