Let the line x-y=4 intersect the circle C:(x-4)^2+(y+3)^2=9 at the points Q and R .

Question

Let the line $x-y=4$ intersect the circle $C:(x-4{)}^{2}+(y+3{)}^{2}=9$ at the points Q and R . If $P(\alpha ,\beta )$ is a point on C such that $PQ=PR$, then $(6\alpha +8\beta {)}^{2}$ is equal to $____$ .

Answer: 18

Accepted Answer

Since $PQ=PR$, the point $P$ must lie on the perpendicular bisector of the chord $QR$.

Now, the points $Q$ and $R$ are the intersection points of the line

$x-y=4$

with the circle

$(x-4{)}^{2}+(y+3{)}^{2}=9.$

So, $QR$ is a chord of the circle lying on the line $x-y=4$.

For a circle, the perpendicular bisector of any chord passes through the centre.

The centre of the circle is

$(4,-3).$

The slope of the line $x-y=4$ is $1$, so the slope of the perpendicular bisector is $-1$.

Hence the perpendicular bisector of chord $QR$ passing through $(4,-3)$ is

$y+3=-1(x-4)$

$y+3=-x+4$

$x+y=1.$

Therefore, point $P(\alpha ,\beta )$ lies on both:

the circle:

$(\alpha -4{)}^{2}+(\beta +3{)}^{2}=9$

the line:

$\alpha +\beta =1.$

Now we need $(6\alpha +8\beta {)}^{2}$.

Using $\alpha +\beta =1$, let

$\beta =1-\alpha .$

Substitute into the circle:

$(\alpha -4{)}^{2}+((1-\alpha )+3{)}^{2}=9$

$(\alpha -4{)}^{2}+(4-\alpha {)}^{2}=9$

Since $(4-\alpha {)}^{2}=(\alpha -4{)}^{2}$,

$2(\alpha -4{)}^{2}=9$

$(\alpha -4{)}^{2}=\frac{9}{2}.$

Now,

$6\alpha +8\beta =6\alpha +8(1-\alpha )=8-2\alpha .$

So,

$(6\alpha +8\beta {)}^{2}=(8-2\alpha {)}^{2}=4(\alpha -4{)}^{2}.$

Using $(\alpha -4{)}^{2}=\frac{9}{2}$,

$(6\alpha +8\beta {)}^{2}=4\cdot \frac{9}{2}=18.$

Hence, the required value is

$18.$

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