Question:
The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is
The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is
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To save time on an exam, you can simply plug the given coordinates $(3,7)$ and $(5,5)$ directly into the four options! The correct equation must evaluate to $0 = 0$ for both points.
Updated On: Jun 1, 2026
- $x^2+y^2+6x-2y+90=0$
- $x^2+y^2-6x-2y-25=0$
- $x^2+y^2-6x+2y-30=0$
- $x^2+y^2+6x+2y-90=0$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We need to find the full expanded equation of a circle. We are given the condition that its center lies on a specific linear equation and that the circumference passes through two exact points.
Step 2: Key Formula or Approach:
Let the center of the circle be $C(h, k)$.
Since it lies on the line $x - 4y = 1$, it must satisfy $h - 4k = 1 \implies h = 4k + 1$.
The distance from the center to any point on the circle is equal to the radius ($r$). Therefore, the distance squared from $C(h,k)$ to point $A(3,7)$ must equal the distance squared from $C(h,k)$ to point $B(5,5)$.
$$(h - 3)^2 + (k - 7)^2 = (h - 5)^2 + (k - 5)^2$$
Step 3: Detailed Explanation:
Substitute $h = 4k + 1$ into the distance equality:
$$(4k + 1 - 3)^2 + (k - 7)^2 = (4k + 1 - 5)^2 + (k - 5)^2$$ $$(4k - 2)^2 + (k - 7)^2 = (4k - 4)^2 + (k - 5)^2$$ Expand all binomials:
$$(16k^2 - 16k + 4) + (k^2 - 14k + 49) = (16k^2 - 32k + 16) + (k^2 - 10k + 25)$$ Combine like terms on both sides:
$$17k^2 - 30k + 53 = 17k^2 - 42k + 41$$ The $17k^2$ terms cancel out perfectly:
$$-30k + 53 = -42k + 41$$ $$12k = -12 \implies k = -1$$ Now, substitute $k = -1$ back to find $h$:
$$h = 4(-1) + 1 = -3$$ The center is $C(-3, -1)$.
Calculate the radius squared ($r^2$) using point $B(5,5)$:
$$r^2 = (-3 - 5)^2 + (-1 - 5)^2 = (-8)^2 + (-6)^2 = 64 + 36 = 100$$ Construct the standard circle equation:
$$(x - h)^2 + (y - k)^2 = r^2$$ $$(x + 3)^2 + (y + 1)^2 = 100$$ Expand to general form:
$$x^2 + 6x + 9 + y^2 + 2y + 1 = 100$$ $$x^2 + y^2 + 6x + 2y - 90 = 0$$
Step 4: Final Answer:
The equation of the circle is $x^2+y^2+6x+2y-90=0$, which corresponds to option (D).
We need to find the full expanded equation of a circle. We are given the condition that its center lies on a specific linear equation and that the circumference passes through two exact points.
Step 2: Key Formula or Approach:
Let the center of the circle be $C(h, k)$.
Since it lies on the line $x - 4y = 1$, it must satisfy $h - 4k = 1 \implies h = 4k + 1$.
The distance from the center to any point on the circle is equal to the radius ($r$). Therefore, the distance squared from $C(h,k)$ to point $A(3,7)$ must equal the distance squared from $C(h,k)$ to point $B(5,5)$.
$$(h - 3)^2 + (k - 7)^2 = (h - 5)^2 + (k - 5)^2$$
Step 3: Detailed Explanation:
Substitute $h = 4k + 1$ into the distance equality:
$$(4k + 1 - 3)^2 + (k - 7)^2 = (4k + 1 - 5)^2 + (k - 5)^2$$ $$(4k - 2)^2 + (k - 7)^2 = (4k - 4)^2 + (k - 5)^2$$ Expand all binomials:
$$(16k^2 - 16k + 4) + (k^2 - 14k + 49) = (16k^2 - 32k + 16) + (k^2 - 10k + 25)$$ Combine like terms on both sides:
$$17k^2 - 30k + 53 = 17k^2 - 42k + 41$$ The $17k^2$ terms cancel out perfectly:
$$-30k + 53 = -42k + 41$$ $$12k = -12 \implies k = -1$$ Now, substitute $k = -1$ back to find $h$:
$$h = 4(-1) + 1 = -3$$ The center is $C(-3, -1)$.
Calculate the radius squared ($r^2$) using point $B(5,5)$:
$$r^2 = (-3 - 5)^2 + (-1 - 5)^2 = (-8)^2 + (-6)^2 = 64 + 36 = 100$$ Construct the standard circle equation:
$$(x - h)^2 + (y - k)^2 = r^2$$ $$(x + 3)^2 + (y + 1)^2 = 100$$ Expand to general form:
$$x^2 + 6x + 9 + y^2 + 2y + 1 = 100$$ $$x^2 + y^2 + 6x + 2y - 90 = 0$$
Step 4: Final Answer:
The equation of the circle is $x^2+y^2+6x+2y-90=0$, which corresponds to option (D).
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