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

Answer: (B)

8

Answer 2

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): $8$