Question

The equation of a chord to the circle $x^{2}+y^{2}=25$ is $x+y-5=0$. The equation of a circle whose diameter is $x+y-5=0$, is}

• A. $x^{2}+y^{2}-6x-5y=0$
• B. $x^{2}+y^{2}-5x-6y=0$
• C. $x^{2}+y^{2}-5x-5y=0$
• D. $x^{2}+y^{2}-6x-6y=0$
• E. $x^{2}+y^{2}-3x-3y=0$

Hint:

Math Tip: You can also use the family of circles formula $S + \lambda L = 0$. For $x^2+y^2-25 + \lambda(x+y-5) = 0$, the center is $(-\lambda/2, -\lambda/2)$. Since the line $x+y-5=0$ is the diameter, the center must lie on it. Substituting the center into the line equation quickly yields $\lambda = -5$.

Answer 1

The Correct Option is C

Concept:
Coordinate Geometry - Circles and Diametric Form.
Step 1: Find the points of intersection of the circle and the chord.
Given the circle equation: $x^2 + y^2 = 25$
Given the chord equation: $x + y - 5 = 0 \implies y = 5 - x$
Substitute $y$ into the circle's equation: $$ x^2 + (5 - x)^2 = 25 $$
Step 2: Solve the quadratic equation for x.
Expand the squared term: $$ x^2 + (25 - 10x + x^2) = 25 $$ $$ 2x^2 - 10x + 25 = 25 $$ $$ 2x^2 - 10x = 0 $$ Factor out $2x$: $$ 2x(x - 5) = 0 $$ This gives two solutions for $x$: $x = 0$ and $x = 5$.
Step 3: Find the corresponding y-coordinates.
Substitute the $x$ values back into $y = 5 - x$:

• When $x = 0$, $y = 5 - 0 = 5$. Point A is $(0, 5)$.
• When $x = 5$, $y = 5 - 5 = 0$. Point B is $(5, 0)$.

These intersection points, $A(0,5)$ and $B(5,0)$, form the endpoints of the diameter for the new circle.
Step 4: Use the diametric form of a circle equation.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is: $$ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 $$
Step 5: Substitute the points and expand.
Substitute $(0, 5)$ and $(5, 0)$: $$ (x - 0)(x - 5) + (y - 5)(y - 0) = 0 $$ $$ x(x - 5) + y(y - 5) = 0 $$ $$ x^2 - 5x + y^2 - 5y = 0 $$ Rearrange to standard form: $$ x^2 + y^2 - 5x - 5y = 0 $$