Aspire Faculty ID #16692 · Topic: CUET 2023 · Just now
CUET 2023

A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2 + y^2 = 16$ and $x^2 + y^2 = 1$. Then

Solution

Let the centre of circle $S$ be $(h,k)$ and radius $r$.

Orthogonality condition with circle $x^2+y^2=1$:
$h^2 + k^2 = r^2 + 1$

Orthogonality with $(x-1)^2+y^2=16$:
$(h-1)^2 + k^2 = r^2 + 16$

Subtracting:
$(h-1)^2 - h^2 = 15$
$h^2 - 2h +1 - h^2 = 15$
$-2h = 14 \Rightarrow h = -7$

Since circle passes through $(0,1)$:
$r^2 = (0+7)^2 + (1-k)^2$

Using $h^2+k^2=r^2+1$:
$49 + k^2 = r^2 + 1$

Solving gives $k=1$.

Centre of $S = (-7,1)$
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