CircleHardBloom L3
Question
If a circle passes through the point $(a, b)$ and cuts the circle $x^2 + y^2 = p^2$ orthogonally, then the equation of the locus of its center is
Options
A.$x^2 + y^2 - 3ax - 4by + (a^2 + b^2 - p^2) = 0$
B.$2ax + 2by - (a^2 - b^2 + p^2) = 0$
C.$x^2 + y^2 - 2ax - 3by + (a^2 - b^2 - p^2) = 0$
D.$2ax + 2by - (a^2 + b^2 + p^2) = 0$
Solution
{"given":"A circle passes through point $(a, b)$ and cuts the circle $x^2 + y^2 = p^2$ orthogonally. We need to find the locus of the center of the required circle.","key_observation":"For two circles to intersect orthogonally, the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ where the circles are $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$. The given circle $x^2 + y^2 = p^2$ has center $(0,0)$ and the required circle must pass through $(a,b)$.","option_analysis":[{"label":"(A)","text":"$x^2 + y^2 - 3ax - 4by + (a^2 + b^2 - p^2) = 0$","verdict":"incorrect","explanation":"This is a quadratic equation in $x$ and $y$ representing a circle, but the coefficients are incorrect. The locus should be linear since we're finding the relationship between center coordinates."},{"label":"(B)","text":"$2ax + 2by - (a^2 - b^2 + p^2) = 0$","verdict":"incorrect","explanation":"This is a linear equation but has incorrect signs. The constant term should be $-(a^2 + b^2 + p^2)$, not $-(a^2 - b^2 + p^2)$."},{"label":"(C)","text":"$x^2 + y^2 - 2ax - 3by + (a^2 - b^2 - p^2) = 0$","verdict":"incorrect","explanation":"This is also a quadratic equation with incorrect coefficients. The coefficient of $y$ should be $-2b$, not $-3b$, and the constant term is wrong."},{"label":"(D)","text":"$2ax + 2by - (a^2 + b^2 + p^2) = 0$","verdict":"correct","explanation":"This is the correct linear equation. Using orthogonality condition and the fact that the circle passes through $(a,b)$, we get $2ax + 2by = a^2 + b^2 + p^2$."}],"answer":"(D)","formula_steps":[]}
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