Question

Equation of a circle passing through the point (1,2) and (3,4) and touching the line $3x+y-3=0$ is

• A. $x^2+y^2-8x-2y+7=0$
• B. $x^2+y^2-8x-2y-7=0$
• C. $x^2+y^2-4x-2y+3=0$
• D. $x^2+y^2-6x-4y+3=0$

Answer 1

Answer: (A)

$x^2+y^2-8x-2y+7=0$

Consider equation of line i.e. $3x+y-3=0$

Put y=0 in above equation, we get,

$3x-3=0$

$\therefore x=1$

Now, put $x=0$ in above equation, we get,

$y-3=0$

$\therefore y=3$

Thus, line passes through $(1,0)$ and $(0,3)$

Refer to the figure. We can draw two circles which can pass through two points $A(1,2)$ and $B(3,4)$ and tangential to the line $3x+y-3=0$

Let center of one of the circles is $C(h,k)$. Thus, AC and BC will be radii of the circle.

By distance formula, we can write as

$d\left(AC\right)=\sqrt{{(h-1)}^2+{(k-2)}^2}$ Equation (1)

Similarly, $d\left(BC\right)=\sqrt{{(h-3)}^2+{(k-4)}^2}$ Equation (2)

But, $d\left(AC\right)=d\left(BC\right)$ (Radii of the same circle)

$\therefore \sqrt{{(h-1)}^2+{(k-2)}^2}=\sqrt{{(h-3)}^2+{(k-4)}^2}$

Squaring both sides, we get,

${(h-1)}^2+{(k-2)}^2={(h-3)}^2+{(k-4)}^2$

$\therefore h^2-2h+1+k^2-4k+4=h^2-6h+9+k^2-8k+16$

On simplifying, we get,

$4h+4k-20=0$

Dividing both sides by 4, we get,

$h+k-5=0$

$\therefore k=5-h$

Putting this value in equation (1), we get,

$r=d\left(AC\right)=\sqrt{{(h-1)}^2+{(5-h-2)}^2}$

$\therefore r=\sqrt{{(h-1)}^2+{(3-h)}^2}$

$\therefore r=\sqrt{h^2-2h+1+9-6h+h^2}$

$\therefore r=\sqrt{{2h}^2-8h+10}$

Now, let us draw a perpendicular from center of circle C to the line $3x+y-3=0$ which is tangential to the circle. Let M be foot of the perpendicular.

$\therefore d\left(CM\right)=r$ (Refer the figure)

$\therefore d\left(CM\right)=\sqrt{{2h}^2-8h+10}$ Equation (3)

Now, Distance between a point $C(h,k)$ and a line $3x+y-3=0$ is calculated by using vector form as,

$d\left(CM\right)=\frac{\left|Equationofline\right|}{\sqrt{{\left(Coefficientofx\right)}^2+{\left(Coefficientofy\right)}^2}}$

$d\left(CM\right)=\frac{|3x+y-3|}{\sqrt{{\left(3\right)}^2+{\left(1\right)}^2}}$

$d\left(CM\right)=\frac{|3x+y-3|}{\sqrt{10}}$

Replace x by h and y by k, we get,

$d\left(CM\right)=\frac{|3h+k-3|}{\sqrt{10}}$

From equation (3), we get,

$\frac{|3h+k-3|}{\sqrt{10}}=\sqrt{{2h}^2-8h+10}$

But, $k=5-h$

$\therefore \frac{3h+5-h-3}{\sqrt{10}}=\sqrt{{2h}^2-8h+10}$

$\therefore \frac{2h+2}{\sqrt{10}}=\sqrt{{2h}^2-8h+10}$

Squaring both sides, we get,

$\frac{{\left(2h+2\right)}^2}{10}={2h}^2-8h+10$

$\therefore 4h^2+8h+4=10\left({2h}^2-8h+10\right)$

$\therefore 4h^2+8h+4={20h}^2-80h+100$

$\therefore {16h}^2-88h+96=0$

Dividing both sides by 8, we get,

${2h}^2-11h+12=0$

$\therefore a=2$, $b=-11$ and $c=12$

$\therefore h=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\therefore h=\frac{-(-11)\pm \sqrt{{(-11)}^2-4\times 2\times 12}}{2\times 2}$

$\therefore h=\frac{11\pm \sqrt{121-96}}{4}$

$\therefore h=\frac{11\pm \sqrt{25}}{4}$

$\therefore h=\frac{11\pm 5}{4}$

We will consider positive root only.

$\therefore h=\frac{16}{4}=4$

$\therefore k=5-4=1$

Thus, coordinates of center of circle are, $C(4,1)$

Thus, From equation (1),

$r=\sqrt{{(4-1)}^2+{(1-2)}^2}$

$\therefore r=\sqrt{9+1}=\sqrt{10}$

Thus, equation of circle will be,

${(x-h)}^2+{(y-k)}^2=r^2$

${(x-4)}^2+{(y-1)}^2={\left(\sqrt{10}\right)}^2$

$\therefore x^2-8x+16+y^2-2y+1=10$

$\therefore x^2+y^2-8x-2y+7=0$

Thus, Answer is option (A)