{"id":25475,"date":"2025-06-19T03:04:14","date_gmt":"2025-06-19T03:04:14","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=25475"},"modified":"2025-06-19T03:04:16","modified_gmt":"2025-06-19T03:04:16","slug":"which-equation-represents-the-circle-described-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/which-equation-represents-the-circle-described-2\/","title":{"rendered":"Which equation represents the circle described"},"content":{"rendered":"\n

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 x2 + y2 \u2013 8x \u2013 6y + 24 = 0.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Answer:<\/h3>\n\n\n\n

(x \u2013 4)\u00b2 + (y \u2013 3)\u00b2 = 4<\/strong><\/p>\n\n\n\n

\n\n\n\n

Explanation:<\/h3>\n\n\n\n

We are given two important pieces of information:<\/p>\n\n\n\n

    \n
  1. The radius<\/strong> of the new circle is 2 units<\/strong>.<\/li>\n\n\n\n
  2. The center<\/strong> is the same as that of another circle whose equation is:
    x2+y2\u22128x\u22126y+24=0x^2 + y^2 – 8x – 6y + 24 = 0x2+y2\u22128x\u22126y+24=0<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Step 1: Find the center of the given circle<\/h4>\n\n\n\n

    We start by rewriting the equation in standard form<\/strong> by completing the square:<\/p>\n\n\n\n

    Given:
    x2+y2\u22128x\u22126y+24=0x^2 + y^2 – 8x – 6y + 24 = 0x2+y2\u22128x\u22126y+24=0<\/p>\n\n\n\n

    Group x and y terms:
    (x2\u22128x)+(y2\u22126y)=\u221224(x^2 – 8x) + (y^2 – 6y) = -24(x2\u22128x)+(y2\u22126y)=\u221224<\/p>\n\n\n\n

    Complete the square:<\/p>\n\n\n\n

      \n
    • For x2\u22128xx^2 – 8xx2\u22128x, take half of 8 (which is 4), square it: 42=164^2 = 1642=16<\/li>\n\n\n\n
    • For y2\u22126yy^2 – 6yy2\u22126y, take half of 6 (which is 3), square it: 32=93^2 = 932=9<\/li>\n<\/ul>\n\n\n\n

      Add 16 and 9 to both sides:
      (x2\u22128x+16)+(y2\u22126y+9)=\u221224+16+9(x^2 – 8x + 16) + (y^2 – 6y + 9) = -24 + 16 + 9(x2\u22128x+16)+(y2\u22126y+9)=\u221224+16+9<\/p>\n\n\n\n

      Simplify:
      (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<\/strong> is at (4, 3)<\/strong> and the radius<\/strong> is 1<\/strong> (from the original circle).<\/p>\n\n\n\n

      \n\n\n\n

      Step 2: Write the new circle’s equation<\/h4>\n\n\n\n

      The new circle has the same center (4, 3)<\/strong> but a radius of 2 units<\/strong>.<\/p>\n\n\n\n

      Using the standard form of a circle:
      (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

      Plug in the values:
      (x\u22124)2+(y\u22123)2=22=4(x – 4)^2 + (y – 3)^2 = 2^2 = 4(x\u22124)2+(y\u22123)2=22=4<\/p>\n\n\n\n

      \n\n\n\n

      Final Answer:<\/h4>\n\n\n\n

      (x \u2013 4)\u00b2 + (y \u2013 3)\u00b2 = 4<\/strong><\/p>\n\n\n\n

      This equation represents a circle centered at (4, 3) with a radius of 2 units.<\/p>\n\n\n\n

      \"\"<\/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 x2 + y2 \u2013 8x \u2013 6y + 24 = 0. The Correct Answer and Explanation is: Correct Answer: (x \u2013 4)\u00b2 + (y \u2013 3)\u00b2 = 4 Explanation: We […]<\/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-25475","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25475","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=25475"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25475\/revisions"}],"predecessor-version":[{"id":25477,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25475\/revisions\/25477"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=25475"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=25475"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=25475"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}