Let the centre of the circle x^2+y^2+2gx+2fy+25=0 be in the first quadrant and lie on the …

Question

Let the centre of the circle ${x}^{2}+{y}^{2}+2gx+2fy+25=0$ be in the first quadrant and lie on the line $2x-y=4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x=1$ is $____$ .

Answer: 80

Accepted Answer

For the circle

${x}^{2}+{y}^{2}+2gx+2fy+25=0,$

the centre is

$(-g,-f)$

and radius is

$r=\sqrt{{g}^{2}+{f}^{2}-25}.$

We are given that the centre lies in the first quadrant, so

$(-g,-f)$

is in the first quadrant.

Also, the centre lies on the line

$2x-y=4.$

So if the centre is $(-g,-f)$, then

$2(-g)-(-f)=4$

that is,

$-2g+f=4.$

Now use the information about the equilateral triangle inscribed in the circle.

For an equilateral triangle of side $a$,

$=\frac{\sqrt{3}}{4}{a}^{2}.$

Given area is

$27\sqrt{3}.$

So,

$\frac{\sqrt{3}}{4}{a}^{2}=27\sqrt{3}$

which gives

${a}^{2}=108\, \Rightarrow \, a=6\sqrt{3}.$

For an equilateral triangle inscribed in a circle, the circumradius is

$R=\frac{a}{\sqrt{3}}.$

Hence,

$r=\frac{6\sqrt{3}}{\sqrt{3}}=6.$

Therefore,

${r}^{2}=36.$

So,

${g}^{2}+{f}^{2}-25=36$

which gives

${g}^{2}+{f}^{2}=61.$

Now solve the system

$-2g+f=4$

and

${g}^{2}+{f}^{2}=61.$

From the first equation,

$f=4+2g.$

Substitute into the second:

${g}^{2}+(4+2g{)}^{2}=61$

${g}^{2}+16+16g+4{g}^{2}=61$

$5{g}^{2}+16g-45=0.$

Solve:

$5{g}^{2}+16g-45=0.$

Discriminant:

$\Delta={16}^{2}-4\cdot 5\cdot (-45)=256+900=1156={34}^{2}.$

Thus,

$g=\frac{-16\pm 34}{10}.$

So,

$g=\frac{18}{10}=\frac{9}{5}\, \, g=\frac{-50}{10}=-5.$

Now centre is $(-g,-f)$ and must be in the first quadrant.

If $g=\frac{9}{5}$, then

$f=4+2\cdot \frac{9}{5}=\frac{38}{5},$

so centre is

$(-\frac{9}{5},-\frac{38}{5}),$

not in first quadrant.

Hence reject.

If $g=-5$, then

$f=4+2(-5)=-6.$

So centre is

$(5,6),$

which is in the first quadrant.

Thus the circle is

${x}^{2}+{y}^{2}-10x-12y+25=0,$

with centre $(5,6)$ and radius $6$.

Now we need the chord cut by the line

$x=1.$

Distance of the line $x=1$ from the centre $(5,6)$ is

$|5-1|=4.$

If a chord is at distance $d$ from the centre in a circle of radius $r$, then chord length is

$2\sqrt{{r}^{2}-{d}^{2}}.$

So here chord length is

$2\sqrt{{6}^{2}-{4}^{2}}=2\sqrt{36-16}=2\sqrt{20}=4\sqrt{5}.$

Therefore, the square of the chord length is

$(4\sqrt{5}{)}^{2}=80.$

Hence, the required answer is

$80.$

Solutions · 1