Question

Let \(\frac{dy}{dx} = \frac{ax-by+a}{bx+cy+a}\), where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is

• A. 10
• B. 8
• C. 7
• D. 5

Answer 1

The Correct Option is B

Let \(\frac{dy}{dx} = \frac{ax-by+a}{bx+cy+a}\)

= \(bx\;dy + cy\; dy + a \;dy = ax \;dx – by\; dx + a\; dx\)

= \(cy\) \(dy\) + \(a\) \(dy\) – \(ax\) \(dx\) – \(a\) \(dx\) + \(b(x dy + y dx)\) = \(0\)

=\(c\) \(∫y\)  \(dy\)+\(a\) \(∫x\) \(dx\)−\(a\) \(∫dx\)+\(b\) \(∫d(xy)=0\)

=\(\frac{cy^2}{2}+ay−\frac{ax^2}{2}−ax+bxy=k\)

= \(ax^2 – cy^2 + 2ax – 2ay – 2bxy = k\)

Above equation is circle

\(⇒ a = –\;c \;and \;b = 0\)

\(ax^2 + ay^2 + 2ax – 2ay = k\)

\(⇒ x^2 + y^2 + 2x – 2y = λ\)

\(\bigg[λ=\frac{k}{a}\bigg]\)

Passes through \((2, 5)\)

\(4 + 25 + 4 – 10 = λ⇒λ = 23\)

Circle \(≡x^2 + y^2 + 2x – 2y – 23 = 0\)

Centre \((–1, 1)\)

\(r=\sqrt{(−1)^2+1^2+23=5}\)

Shortest distance of \((11, 6)\)

=\(\sqrt{12^2+5^2}−5\)

= \(13 – 5\)

= \(8\)

Hence, the correct option is (B): \(8\)