
The locus of the centre of the circle which cuts off intercepts of length \[2a\] and \[2b\] from \[x\] -axis and \[y\]-axis, respectively, is?
A. \[x + y = a + b\]
B. \[{x^2} - {y^2} = {a^2} - {b^2}\]
C. \[{x^2} - {y^2} = {a^2} - {b^2}\]
D. \[{x^2} + {y^2} = {a^2} - {b^2}\]
Answer
163.8k+ views
Hint: In this problem to find the locus of the centre of the circle which cuts off intercepts of length \[2a\] and \[2b\]. To do so, you need to use the squaring and subtracting method. This involves making various smaller circles until you have found all of the intersections between the original Circle and your new Circles. Once you've done that, divide each intersection by its respective radius (\[2a\] for point \[A\] and \[2b\] for point \[B\]) to get their respective lengths. Then add these together to create your final answer.
Complete step-by-step solution :
The locus of the center of a circle is typically defined as the point that lies at the intersection of the radius and all straight lines drawn from any point on the circumference to that same point.
Given the information provided, the locus of the centre of the circle which cuts off intercepts of length \[2a\] and \[2b\].
We know that
\[X\] intercepts \[2\sqrt {{g^2} - c} = 2a\] …………(i)
\[Y\] intercepts \[2\sqrt {{f^2} - c} = 2b\] …………(ii)
By squaring and subtracting (i) and (ii) we get,
\[{\left. {2\sqrt {{g^2} - c} } \right)^2} - {\left( {2\sqrt {{f^2} - c} } \right)^2} = 4{a^2} - 4{b^2}\]
\[4\left( {{g^2} - c} \right) - 4\left( {{f^2} - c} \right) = 4{a^2} - 4{b^2}\]
\[{g^2} - c - {f^2} + c = {a^2} - {b^2}\]
\[{g^2} - {f^2} = {a^2} - {b^2}\]
As a result, the locus is \[{x^2} - {y^2} = {a^2} - {b^2}\].
Hence, a circle's centre that cuts off intercepts of 2a and 2b from x-axis and y-axis is
\[{x^2} - {y^2} = {a^2} - {b^2}\].
So, the correct answer is option B.
Note :
A student might make a mistake when trying to solve this equation by assuming that the locus is a point. In reality, the locus is more likely to be a curve. If they don't understand the concept of intercepts, Intercepts are points where the slope of a linear function changes sign, which means that they become positive or negative depending on the direction that you look at them.
Complete step-by-step solution :
The locus of the center of a circle is typically defined as the point that lies at the intersection of the radius and all straight lines drawn from any point on the circumference to that same point.
Given the information provided, the locus of the centre of the circle which cuts off intercepts of length \[2a\] and \[2b\].
We know that
\[X\] intercepts \[2\sqrt {{g^2} - c} = 2a\] …………(i)
\[Y\] intercepts \[2\sqrt {{f^2} - c} = 2b\] …………(ii)
By squaring and subtracting (i) and (ii) we get,
\[{\left. {2\sqrt {{g^2} - c} } \right)^2} - {\left( {2\sqrt {{f^2} - c} } \right)^2} = 4{a^2} - 4{b^2}\]
\[4\left( {{g^2} - c} \right) - 4\left( {{f^2} - c} \right) = 4{a^2} - 4{b^2}\]
\[{g^2} - c - {f^2} + c = {a^2} - {b^2}\]
\[{g^2} - {f^2} = {a^2} - {b^2}\]
As a result, the locus is \[{x^2} - {y^2} = {a^2} - {b^2}\].
Hence, a circle's centre that cuts off intercepts of 2a and 2b from x-axis and y-axis is
\[{x^2} - {y^2} = {a^2} - {b^2}\].
So, the correct answer is option B.
Note :
A student might make a mistake when trying to solve this equation by assuming that the locus is a point. In reality, the locus is more likely to be a curve. If they don't understand the concept of intercepts, Intercepts are points where the slope of a linear function changes sign, which means that they become positive or negative depending on the direction that you look at them.
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