Question

The equations of the circles which touch the lines $3x-4y+1=0$ and $4x+3y-7=0$ and pass through $(2,3)$ are

• A. $(x-\frac{6}{5}{)}^2+(y-\frac{12}{5}{)}^2=1,(x-2{)}^2+(y-8{)}^2=25$
• B. $(x+\frac{6}{5}{)}^2+(y+\frac{12}{6}{)}^2=1,(x-2{)}^2+(y-8{)}^2=25$
• C. $(x+\frac{6}{5}{)}^2+(y+\frac{12}{6}{)}^2=1,(x+2{)}^2+(y+8{)}^2=25$
• D. . $(x-\frac{6}{5}{)}^2+(y-\frac{12}{6}{)}^2=1,(x+2{)}^2+(y+8{)}^2=25$

Answer 1

The correct option is A $(x-\frac{6}{5}{)}^2+(y-\frac{12}{5}{)}^2=1,(x-2{)}^2+(y-8{)}^2=25$

$3x-4y+1=0$ and $4x+3y-7=0$ are perpendicular tangents to a circle.

Let $c(h,k)$ is a center of the circle.

Then, $r=\left|\frac{3h-4k+1}{5}\right|=\left|\frac{4h+3k-7}{5}\right|$

$3h-4k+1=4h+3k-7$ and $3h-4k+1=-\left(4h+3k-7\right)$

From solving above we get, $h+7k-8=0$ -----(1) and $7h-7k-6=0$ -----(2)

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

From these three equations,

$h=\frac{6}{5}$ or $h=2$

and $k=\frac{12}{5}$ or $k=8$

So, $r=1$ or $r=5$

Then ${(x-\frac{6}{5})}^2+{(y-\frac{12}{5})}^2=1$

and $(x-2{)}^2+(y-8{)}^2=25$ are two circles possible.