A circle with centre (2,3) and radius 4 intersects the line x+y=3 at the points P and Q. If the tangents at P and Q intersect at S(α,β), then 4α−7β is equal to \_\_\_\_\_\_\_\_.
Answer: 11
Solutions · 1
Official solution
Circle: x2+y2−4x−6y−3=0. The line x+y=3 is the chord of contact from S(α,β), so the chord of contact equation
\[\alpha x+\beta y-2(x+\alpha)-3(y+\beta)-3=0\ \Longrightarrow\ (\alpha-2)x+(\beta-3)y-(2\alpha+3\beta+3)=0\]
must be proportional to x+y−3=0. Comparing coefficients: α−2=β−3 and 3(α−2)=2α+3β+3.
From these: α=β−1 and α=3β+9, giving β−1=3β+9⇒β=−5,α=−6.
\[4\alpha-7\beta=4(-6)-7(-5)=-24+35=11.\]