Question

The equation of circle described on the chord $3x+y+5=0$ of the circle $x^2+y^2=16$ as diameter is:

• A. $x^2+y^2+3x+y-11=0$
• B. $x^2+y^2+3x+y+1=0$
• C. $x^2+y^2+3x+y-2=0$
• D. $x^2+y^2+3x+y-22=0$

Answer 1

Answer: (A)

$x^2+y^2+3x+y-11=0$

For a circle with center (g,f) and radius r, equation is,

$\to (x-g{)}^2+(y-f{)}^2=r^2$

So, radius of circle $x^2+y^2=16$ is 4.

Let (h,k) be the foot of perpendicular from center of circle $x^2+y^2=16$ on line $3x+4y+5=0$.

Product of slopes of two perpendicular lines is $-1$.

So, product of slope of $3x+y+5=0$ and perpendicular on it is $-1$.

$\to -3\times \frac{k}{h}=-1$

$\to 3k=h$

Also, (h,k) lies on line 3x + 4y + 5 = 0

$\to 3h+k+5=0$

$\to 10k+5=0$

$\to k=\frac{-1}{2}$

Hence, $(h,k)=(\frac{-3}{2},\frac{-1}{2})$

Perperndicular from center of circle to its chord bisects chord, so (h,k) is center of new circle.

No, as per figure,

AB is radius of new circle.

To find AB let's first find AC,

AC

= Perperndicular distance from origin to (h,K)

$=\sqrt{(\frac{-1}{2}{)}^2+(\frac{-3}{2}{)}^2}$

$=\sqrt{\frac{5}{2}}$

Now,

AB =

$\sqrt{4^2-(\sqrt{\frac{5}{2}}}{)}^2$

=$\sqrt{16-\frac{5}{2}}$

$=3\sqrt{\frac{3}{2}}$

Now, equation of new circle,

$\to (x+\frac{3}{2}{)}^2+(y+\frac{1}{2}{)}^2=\frac{27}{2}$

$\to x^2+y^2+3x+y-11=0$

Hence, (A) is the correct answer.