To find the correct equation of the circle, follow these steps:<\/p>\n\n\n\n
Step 1: Identify the center of the given circle<\/h3>\n\n\n\n
We are given a circle with the equation: x2+y2\u22128x\u22126y+24=0x^2 + y^2 – 8x – 6y + 24 = 0x2+y2\u22128x\u22126y+24=0<\/p>\n\n\n\n
To find its center, complete the square.<\/p>\n\n\n\n
Group the terms: (x2\u22128x)+(y2\u22126y)+24=0(x^2 – 8x) + (y^2 – 6y) + 24 = 0(x2\u22128x)+(y2\u22126y)+24=0<\/p>\n\n\n\n
Complete the square for both groups:<\/p>\n\n\n\n
- \n
- For x2\u22128xx^2 – 8xx2\u22128x, add and subtract 161616 (82)2(\\frac{8}{2})^2(28\u200b)2<\/li>\n\n\n\n
- For y2\u22126yy^2 – 6yy2\u22126y, add and subtract 999 (62)2(\\frac{6}{2})^2(26\u200b)2<\/li>\n<\/ul>\n\n\n\n
(x2\u22128x+16)+(y2\u22126y+9)+24\u221216\u22129=0(x^2 – 8x + 16) + (y^2 – 6y + 9) + 24 – 16 – 9 = 0(x2\u22128x+16)+(y2\u22126y+9)+24\u221216\u22129=0 (x\u22124)2+(y\u22123)2\u22121=0(x – 4)^2 + (y – 3)^2 – 1 = 0(x\u22124)2+(y\u22123)2\u22121=0 (x\u22124)2+(y\u22123)2=1(x – 4)^2 + (y – 3)^2 = 1(x\u22124)2+(y\u22123)2=1<\/p>\n\n\n\n
So, the center of this circle is (4,3)(4, 3)(4,3)<\/p>\n\n\n\n
\n\n\n\nStep 2: Use the new radius<\/h3>\n\n\n\n
Now, we are told to find a new circle with the same center<\/strong> (4,3)(4, 3)(4,3) and radius of 2 units<\/strong><\/p>\n\n\n\n
The standard form of a circle’s equation is: (x\u2212h)2+(y\u2212k)2=r2(x – h)^2 + (y – k)^2 = r^2(x\u2212h)2+(y\u2212k)2=r2<\/p>\n\n\n\n
Where (h,k)(h, k)(h,k) is the center and rrr is the radius.<\/p>\n\n\n\n
So the equation becomes: (x\u22124)2+(y\u22123)2=22(x – 4)^2 + (y – 3)^2 = 2^2(x\u22124)2+(y\u22123)2=22<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
c. (x\u22124)2+(y\u22123)2=22(x – 4)^2 + (y – 3)^2 = 2^2(x\u22124)2+(y\u22123)2=22<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n
This equation correctly reflects a circle with radius 2 and center (4,3)(4, 3)(4,3). It matches choice c<\/strong>, which shows the squared radius rather than the simplified number. This form is acceptable and commonly used to emphasize the actual radius. Choices a, b, and d either have the wrong center or do not correctly express the squared radius.<\/p>\n\n\n\n
![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-33.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Which equation represents the circle described? The radius is 2 units. The center is the same as the center of a circle whose equation is x^2 + y^2 – 8x – 6y + 24 = 0. a. (x + 4)^2 + (y + 3)^2 = 2 b. (x – 4)^2 + (y – 3)^2 = […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-25470","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25470","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=25470"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25470\/revisions"}],"predecessor-version":[{"id":25473,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25470\/revisions\/25473"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=25470"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=25470"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=25470"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}