Question

Assertion :If the point on a circle nearest to the point $P(2,1)$ is at $4$ unit distance and the farthest is $(6,5)$, then the equation of the circle is $(x-6)(x-2-2\sqrt{2})+(y-5)(y-1-2\sqrt{2})=0$ Reason: The equation of a circle having end points of the diameter as $(x_1,y_1)$ and $(x_2,y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Given $P=(2,1)$ and $PA=4$ where $A$ is the nearest point and $B=(6,5)$ , $B$ is the farthest point.

$P,A$ and $B$ all lie on the same line.

$PB=\sqrt{4^2+4^2}=4\sqrt{2}⟹AB=4\sqrt{2}-4$

$A$ divides $P$ and $B$ in the ratio$4:4\sqrt{2}–4$

i.e $1:(\sqrt{2}–1)$ internally.

$A=\frac{(\sqrt{2}-1)P+1.B}{(\sqrt{2}–1)+1}$

$=(\frac{(2\sqrt{2}–2)+6}{\sqrt{2}},\frac{(\sqrt{2}-1)+5}{\sqrt{2}})$

$=(2+2\sqrt{2},1+2\sqrt{2})$

Equation of circle having A and B as ends of diameter is $(x–6)(x–2-2\sqrt{2})+(y–5)(y–1–2\sqrt{2})$