Question

The length of tangents from P(1, -1) and Q(3,3) to a circle are $\sqrt{2}and\sqrt{6}$ respectively, then the length of tangent from R(-2,-7) to the same circle is

• A. $\sqrt{41}$
• B. $\sqrt{51}$
• C. $\sqrt{6}$
• D. $\sqrt{71}$

Answer 1

The correct option is D $\sqrt{71}$

Given, P (1,-1) and Q (3,3) to a circle are $\sqrt{2}\&\sqrt{6}$

then,

PA=$\sqrt{2}$

$P{A}^2$=2

Now,length of tengent

$\begin{array}{c}\sqrt{{(x+g)}^2+{(y+f)}^2-g^2-f^2+c}\\ Assumethat\\ circleequation.........\\ x^2+y^2+2gy+2fy+c=0\\ then,\\ PA=\sqrt{{(1+g)}^2+{(-1+f)}^2-g^2-f^2+c}\\ 2=1+g^2+2g+1+f^2-2f-g^2-f^2+c\\ 2=2+2g-2f+c\\ \therefore 2g-2f+c=0.-----------\left(1\right)\\ then,\\ QB=\sqrt{6}\\ Q{B}^2=6\\ lengthoftengent......\\ QB=\sqrt{{(3+g)}^2+{(3+f)}^2-g^2-f^2+c}\\ 6=9+g^2+6g+9+f^2+6f-g^2-f^2+c\\ 6=18+6g+6f+c\\ 6g+6f+c=-12---------\left(2\right)\\ suppose:ciscontant,\\ g=-1-\frac{c}{3},f=-1+\frac{c}{6}\end{array}$

$\begin{array}{c}And,\\ RC=\sqrt{{(-2+g)}^2+{(-7+f)}^2-g^2-f^2+c}\\ =\sqrt{4+g^2-4g+49+f^2-14f-g^2-f^2+c}\\ R{C}^2=53-4(-1-\frac{c}{3})-14(-1+\frac{c}{6})+c\\ =53+4+\frac{4c}{3}+14-\frac{7c}{3}+c\\ R{C}^2=71\\ \therefore RC=\sqrt{71}\\ So,thatthecorrectoptionisD\end{array}$