Question

The equations of circles with radius $3$ units and touching the circle $x^2+y^2-2x-4y-20=0$ at $(5,5)$ is/are

• A. $(5x-16{)}^2+(5y-13{)}^2=225$
• B. $(5x-13{)}^2+(5y-16{)}^2=225$
• C. $(5x-34{)}^2+(5y-37{)}^2=225$
• D. $(5x-37{)}^2+(5y-34{)}^2=225$

Answer 1

Answer: (B), (D)

$(5x-37{)}^2+(5y-34{)}^2=225$


Given circle is $x^2+y^2-2x-4y-20=0$
$C=(1,2),r=5$

As seen from figure, there are $2$ possible circles of $3$ units,
Now, slope of the line joining centres
$\tan \theta =\frac{5-2}{5-1}=\frac{3}{4}\Rightarrow \cos \theta =\frac{4}{5},\sin \theta =\frac{3}{5}$

Using parametric form of line
The coordinates of $C_1$ and $C_2$ are
$=(5\pm 3\cos \theta ,5\pm \sin \theta )=(5\pm 3\left(\frac{4}{5}\right),5\pm 3\left(\frac{3}{5}\right))\Rightarrow C_1=(\frac{13}{5},\frac{16}{5}),C_2=(\frac{37}{5},\frac{34}{5})$

$\therefore$ The required equations of the circles are
$(x-\frac{13}{5})+(y-\frac{16}{5})=9\Rightarrow (5x-13{)}^2+(5y-16{)}^2=225(x-\frac{37}{5})+(y-\frac{34}{5})=9\Rightarrow (5x-37{)}^2+(5y-34{)}^2=225$




Alternate solution:
Given circle is $x^2+y^2-2x-4y-20=0$
$C=(1,2),r=5$
Let the centre of the circle be $(h,k)$
Now,
$\frac{k-5}{h-5}=\frac{5-2}{5-1}\Rightarrow \frac{k-5}{h-5}=\frac{3}{4}\cdots \left(1\right)$
And
$(h-5{)}^2+(k-5{)}^2=3^2$
Using equation $\left(1\right)$, we get
$\Rightarrow \frac{25(h-5{)}^2}{16}=9\Rightarrow h-5=\pm \frac{12}{5}\Rightarrow h=\frac{13}{5},\frac{37}{5}\Rightarrow k=\frac{16}{5},\frac{34}{5}$

So, the equation of circles are
$(x-\frac{13}{5})+(y-\frac{16}{5})=9\Rightarrow (5x-13{)}^2+(5y-16{)}^2=225(x-\frac{37}{5})+(y-\frac{34}{5})=9\Rightarrow (5x-37{)}^2+(5y-34{)}^2=225$